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Lets examine a typical GR metric:

$$ds^2=g_{00}dt^2-g_{11}dx^2-g_{22}dy^2-g_{33}dz^2$$

The "d" going with ds has its correct meaning when the path is specified with respect to a one dimensional manifold (remembering that ds is the proper time interval which will depend on path).

The physical distance (spatial) between two points along the x-axis between the points A and B is given by: $\int\sqrt{g_{11}}dx$ from A to B and not by $\int dx$ in curved space.

Infinitesimal separation between points on the x axis are given by $g_{11}dx$ and not by $dx$.

Now in Maxwell's equations in the covariant form we have quantities like delta-x,delta-y etc which are meaningful only in the Euclidean (rather in the flat space-time Lorentzian) context.But Maxwell's equations in the covariant form refer to curved space time (with respect to strongly curved spacetime also). Are these quantities ($\partial x$,$\partial y $ etc.) expected to retain their physical significance in curved space-time except that they remind us of an Euclidean background?

Better we could write (locally):

$$ds^2=dT^2-dX^2-dY^2-dZ^2$$

Where,

$$\begin{align}dT&=\sqrt{g_{00}}dt \\ dX &= \sqrt{g_{11}}dx \\ dY &= \sqrt{g_{22}}dy \\ dZ &= \sqrt{g_{33}}dz\end{align}$$

(The "d" going with T,X,Yand Z is as justified as the d going with s.)

Locally we have,

$$ds^2=dT^2-dX^2-dY^2-dZ^2$$

Therefore locally we have the same form of Maxwell's equations-- Maxwell's equations in the traditional form!

Though the form of Maxwell’s equations (traditional form being referred to here) remains unchanged locally, the values of the individual variables may change, preserving the traditional form of Maxwell’s equations in the local inertial frames.

We may consider a pair of local labels $x$ and $x+dx$. The physical distance between them along the x-axis is $g_{11}dx$. If the metric changes, say due to the advance of a heavy mass or a high density mass distribution, the physical intervals $dX$,$dY$ etc will change. To preserve the form of the equation the values of $E$,$B$,$j$ etc should also change.

So gravity can change the magnitudes of $E$,$B$ etc. (and of course their directions). If one thinks in the cosmological direction the curvature of space-time was very strong in the remote past and gradually it weakened casting a heavy influence on the values of the electric and the magnetic fields.

Query: How are they taking care of this in the LHC experiments in tracing the past?

[Incidentally the quantities x,y,z etc are simply labels in the curved spacetime context. dx should correspond to some "Euclidean memory"]

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The LHC isn't "tracing the past", they're trying to simulate it at a smaller scale. "Tracing the past" has already been done with cosmic background radiation. Neither is the LHC simulating the entire situation. They're just doing high-energy collisions, and hoping for certain indications to come from it. They are not simulating highly curved spacetime. –  Manishearth Feb 9 '12 at 1:40
    
The amount of deviation from flat spacetime situation is relevant to the context.Here is a link that seems to indicate sufficient curvature changes for $B$ or $E$ to respond press.web.cern.ch/public/en/LHC/LHC-en.html –  Anamitra Palit Feb 9 '12 at 5:50
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The cern link you give does not claim to reproduce the gravity of Big Bang. It just says that it reproduces the average energies of the time after inflation stopped, when particles could coalesce out of the energy glob that was the universe. In en.wikipedia.org/wiki/Big_Bang one gets the timeline. The gravity at LHC is the usual earth one, and any effect is infinitesimally small with respect to measurement capabilities. –  anna v Feb 9 '12 at 12:35
    
As you suggest, CERN is trying to replicate the past without the Gravity---and it expects to get correct inferences!Even a miniature simulation showing sufficient deviation[within calculation or estimation] from flat space time could have been more reliable. With such deviations one should have an estimate of how E and B respond to Gravity changes.So far as you comments are concerned LHC ignores any Unification or interrelationship between Gravitation and EM[and other interactions] –  Anamitra Palit Feb 9 '12 at 14:38
    
How should E and B respond to changes in Gravity?I have tried to suggest methods in the following article: Link: eurojournals.com/ejsr_41_2_06.pdf –  Anamitra Palit Feb 9 '12 at 15:12
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2 Answers

Query: How are they taking care of this in the LHC experiments in tracing the past?

LHC experiments are searching for new particles and interactions at an energy scale which describes a time slice of the evolution of matter after the Big Bang. You may say that it is searching for knowledge of how matter evolved quantitatively in the lab. Much higher energies are displayed by cosmic rays and studied with long arrays of detectors in other experiments. Experiments are studying and searching for constants of motion : masses, spins, decay chains ...

E and B are not constants of motion and it is not surprising that they would not be so when GR is included. The values of the constants of motion searched and the correlations studied are not affected by any E and B dependence . That is why we study constants of motion, instead of recording the changes in the electric field of the proton or the magnetic field of the neutron or .... Invariants is the game.

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$E$ and $B$ are not constants of motion--that's true . But if you consider the energy of a closed system--this energy should remain constant. If E and B change the energy density at each point of a closed system would change--and the total energy is likely to change. –  Anamitra Palit Feb 9 '12 at 5:56
    
If you consider the entire universe you may think in terms of inter-coversion between EM energy and other forms of energy --everything occurring within the universe of course –  Anamitra Palit Feb 9 '12 at 6:00
    
Changes in EM energy is linked inseparably with the expansion of the universe--reflected in the changes in spacetime curvatutre.It[EM energy] looks like an important source powering the expansion of the universe itself! –  Anamitra Palit Feb 9 '12 at 6:29
    
@Anamitra: this has nothing to do with the experiments of the LHC, or the price of tea in China either. LHC is not studying the total energy of the cosmos, but masses and symmetries arising from organizing such masses. How this data is fitted in a Big Bang model or any other cosmological model is the responsibility of the models. In any case measurements in LHC will not affect whether energy is conserved in cosmic scales or not ( I have read here that energy conservation has no meaning in strong GR space distortions) –  anna v Feb 9 '12 at 6:50
    
You may think of a system comprising the particles involved in the collision. My points become relevant with the increase in curvature as the collisions take place. –  Anamitra Palit Feb 9 '12 at 7:00
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A lot of the things you wrote are actually somewhat close to some modern notations and methods used in GR. However, first there are a few things that need to be mentioned. First of all, your metric is far from typical, The majority of Einsteinian metrics are not diagonal as you first mentioned. There really should be 10 terms in that sum, not four... Secondly, I would like to alert you to quantities in tensor analysis known as "tensor densities". These quantities come "weighted" with modified coefficients of the metric. There are scalar densities, vector densities etc... and slightly resemble your redefined terms in the metric, though their construction is a bit more rigorous. Finally, Maxwell's equations can certainly be written in a form that jives perfectly well with Einsteinian manifolds. Two of them are $$ \nabla_{\mu}F^{\mu \nu}=J^{\nu} $$ They just look the same but instead we replace the partial derivative with the covariant derivative. These are suited for use in an arbitrary Einsteinian manifold.

I hope this helps,

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You equation contains quantities like $\partial x$,$\partial y$ etc that correspond to a Euclidean[flat spacetime ]background----a frame preferred by commonsense intuition.A person working with his experiments in strong curved space should consider a "hypothetical " flat space in his interpretations/experiments.Better he/she considers the technique given in the question –  Anamitra Palit Feb 9 '12 at 6:09
    
That isn't true, the equation holds for any curved spacetime. In $\nabla_\mu = \partial_\mu +\Gamma^{\alpha}_{\mu\nu}$ the connection carries information about how to change across a curved manifold. More over, this equation is general, it doesn't even depend on the type of curved manifold, it holds for all Einstinian manifolds; including flat spacetime, among others. –  kηives Feb 9 '12 at 15:35
    
@knives: The Christoffel symbols vanish in the local inertial frames. The best/easiest way to choose local inertial frames is to consider the "physical separations" described in the question. –  Anamitra Palit Feb 9 '12 at 15:52
    
I guess I have to say then that I don't get your question... either we are in a local inertial frame or we aren't. Either there is a non-vanishing first derivative or there isn't, if it's vanishing then Maxwell's equations are just fine, and if not they are just fine too... –  kηives Feb 9 '12 at 20:32
    
Interestingly, metrics for linearized gravity for slow-moving stuff with symmetric (or smalL) stresses are always diagonal, so this is a good approximation. You can see this by expanding isotropic Schwarzschild to linear order. –  Ron Maimon May 21 '12 at 21:55
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