Physics Forums - Special & General Relativity

Thomas Precession, Angular Momentum, and Rotating Reference Frames

Izzhov — Thu, 23 May 2013 23:37:54 GMT

If any of you have the Third Edition of Classical Electrodynamics by John David Jackson, turn to section 11.8, as that's where I'm getting all this from. If not, you should still be able to follow along.

In said section, Jackson gives us this equation that relates any physical vector G in a rotating vs. non-rotating reference frame:

[itex]\left(\frac{d\mathbf{G}}{dt}\right)_{nonrot} = \left(\frac{d\mathbf{G}}{dt}\right)_{rest frame} + \boldsymbol{\omega}_T \times \mathbf{G}[/itex]

where

[itex]\boldsymbol{\omega}_T = \frac{\gamma^2}{\gamma+1}\frac{\mathbf{a}\times\mathbf{v}}{c^2}[/itex]

"where a is the acceleration in the laboratory frame," according to the textbook. Also, gamma is defined using v, the velocity of the particle in the lab frame.

Ok. So I decided to check this by setting G = x, the position vector, for a particle that is undergoing circular motion in the laboratory frame. So we have

[itex]\left(\frac{d\mathbf{x}}{dt}\right)_{nonrot} = \mathbf{v}[/itex]

and

[itex]\left(\frac{d\mathbf{x}}{dt}\right)_{rest frame} = 0[/itex] because the particle doesn't have any velocity in its own frame.

So far so good. Now, this implies that [itex]\boldsymbol{\omega}_T \times \mathbf{x} = \mathbf{v}[/itex]. So if we can get this result from the definition of [itex]\boldsymbol{\omega}_T[/itex], we're golden. However, if you use the fact that [itex]|a| = \frac{v^2}{|x|}[/itex] for circular motion as well as the fact that a is perpendicular to v, and that a is parallel (really antiparallel) to x, and carefully apply the right hand rule, you'll find, after the algebraic dust settles, that

[itex]\boldsymbol{\omega}_T \times \mathbf{G} = (1-\gamma)\mathbf{v}[/itex]

So this is definitely a contradiction. Because it implies that [itex]\mathbf{v} = (1-\gamma)\mathbf{v}[/itex]. Can anyone tell me where this went horribly horribly wrong? I worked on this with my professor for two hours today and we couldn't figure it out.

Minkowski diagram, what does observer see.

Myslius — Thu, 23 May 2013 22:53:04 GMT

Suppose we have a spacetime diagram like this:

Red lines indicating light travel from the moving object to the observer.

Object is moving at the speed of 0.8c. At this speed we have:
Lorentz factor 1/√(1 - v²/c²)=1/0.6=1.66(6)
Relativistic Doppler effect √((1 + β) / (1 - β)) = 3

My question is what does the observer see?
According to this diagram At time t=3 (let's say year) observer sees object 1 year old. At t=6, 2 years old. 6/2=3 and we get Doppler effect.

As far as i know moving at 0.8c speed in no matter what direction relative to the observer results in time ticking at the speed 0.6 (or 1.66(6) slower).

Suppose at t=6 (t'=2) object starts to move towards the observer (same speed).
At t=6, object should arrive to the observer, and be 3.6 years old.

But how's that possible?
How could it be that observer sees the object at time t=6 at both places: arrived, and just starting to making a turn?

I don't understand.

Einstein notion of time and the oscillation of the cesium atom

azoulay — Thu, 23 May 2013 17:28:25 GMT

I just read the thread entitled: "How did Einstein Define Time" and I'm very confused.

At school, I was taught that time was an abstract representation of movement meaning that the word "time" can only be used to represent movements.
For example, when earth has completed a cycle around the Sun, that is called "one year". So in this example we see that the "one year" (concept of time) represents a movement (the cycle of Earth going around the Sun).
So the definition of time for me has always been that it's an abstract representation of some movement and nothing else.

Now, in the Hafele�Keating experiment where Cesium atomic clocks where used to test Einstein's theory of relativity, they state that the clocks going eastward in the airplane jets "lost time" and the clocks going westward "gain time". These conclusions (of clocks gaining and losing time) can only be considered in regards to a specific definition of the notion of time which I think Einstein has lacked to provide. One thing is for sure, considering the definition of time that was taught to me, it is absurd to think that a clock can gain or lose time, it just makes no sense at all.

What does makes sense tough is that while the air plane jets goes eastward, the speed of the cesium atoms oscillations slows down and when the air plane jets goes westward, the speed of the cesium atoms oscillations increases. But that's it !!!! Nothing else can be said about that experiment. The jet planes moving eastward or westward has absolutely NO direct effect on the clock, they only affect the speed of oscillation of the cesium atoms and it is that speed of oscillation of the cesium atoms that's speeding or slowing the clock !!! So the "time" shown on the clock is totally dependant on the speed of oscillation of the cesium atom.

For Einstein to say that the clocks are "losing or gaining time" relative to those clocks travelling at some speed is to make a direct relationship between "time" and the speed of the oscillation of the cesium atoms. So why bother confusing people using the word "time" instead of simply saying things as they are: "it is the speed of oscillation of the cesium atom that is relative" ?

Time is NOT relative to anything unless your definition of time is : "the speed of oscillation of the cesium atom" !!!

I think that Einstein biggest problem was to talk about time without ever giving a specific definition of what he considered time to be. He gave the definition of: "time of an event" but that's different from the notion of "time" alone by itself.

Does my above explanations makes sense to anyone ?

regards,
jonathan

Time synchronization in arbitrary reference frames.

center o bass — Thu, 23 May 2013 14:42:37 GMT

I'm having a bit of trouble with digesting an argument for which the conclusion is that it's not possible, in general, to Einstein synchronize clocks around closed curves. The argument goes as follows;

(Beginning)
Consider a four-velocity field ##e_{\hat{0}}## of the reference particles in a reference frame R.
In an arbitrary basis ##\{e_\mu\}## where we choose ##e_0## parallel to ##e_{\hat{0}}## the vectors ##e_i## need not be parallel to ##e_0##. We thus introduce

$$ e_{\perp i} = e_i - \frac{e_i \cdot e_0}{e_0 \cdot e_0} e_0 $$

and define

$$ \gamma_{ij} = e_{\perp i}\cdot e_{\perp j} = g_{ij} - \frac{g_{i0}g_{j0}}{g_00}$$

where ##g_{\mu \nu}## is the components of the metric tensor corresponding to the original general basis. Written in terms of the time orthogonal basis ##\{e_{\hat 0}, e_{\perp,i}\}## then we get

$$ds^2 = - d\hat t^2 + \gamma_{ij}dx^i dx^j.$$

We can thus write

$$ d\hat t^2 = \gamma_{ij}dx^i dx^j - ds^2 = \gamma_{ij}dx^i dx^j - g_{\mu \nu} dx^\mu dx^\nu = [(-g_{00})^{1/2}(dx^0 + \frac{g_{i0}}{g_{00}} dx^i)]^2.$$

Therefore ##d\hat t=0## corresponds to

$$ dx^0 = - \frac{g_{i0}}{g_{00}} dx^i$$

which is not a perfect differential. Thus the line integral of the coordinate time around a closed curve is in general not zero. The author then concludes that since ##d\hat t = 0## corresponds to simultaneity on Einstein synchronized clocks it is not in general (##g_{i0} \neq 0##) possible to Einstein synchronize clocks around closed curves.
(End)

Questions:
(1) Why does ##d\hat t = 0## correspond to Einstein synchronization of clocks?
(2) Is so bad that the line integral of the coordinate time is different from zero? Does this generally implies something other than a bad choice of time coordinate?

Assigning energy (and maybe mometum) to part of a system in GR.

pervect — Wed, 22 May 2013 23:55:47 GMT

In the case of Komar mass, we can express the mass as an integral of ##\rho + 3P##, so we can meaningful divide the total mass (or energy) of a system into the contribution due to each part, just by integrating over that spatial part of the system.

What happens if we try to do this with other definitions of mass, say the ADM or Bondi mass? My overall impression is that it can't be done, but I don't have a specific reference for this, so I want to be cautious about saying it cant be done.

I suppose I'm open to general ways of partitioning the mass, and not just my suggested approach of integrating some (pseudo) tensor of some sort over a spatial region.

One obstacle that comes to mind with psuedotensors is the obvious issue of the gauge degree of freedom affecting the subdivision process. But this seems lacking as a proof of impossibility, at least without an example illustrating different "partitioning".

Is there always the same "amount" of spacetime curvature in the uni.?

49ers2013Champ — Wed, 22 May 2013 22:33:22 GMT

Universe is what I meant by uni.

Okay, if matter and energy cannot be created or destroyed, and since they are what causes spacetime to curve, does that mean there will always be the same amount of spacetime curvature occuring in the universe?

I understand that large marterial bodies curve spacetime more than the small ones, but regardless of their size or distribution, is there always the same amount of gravity occuring in the universe?

I am sure "amount" is probably not the best word, but I think you will understand what I'm asking.

Ehrenfest's paradox

center o bass — Wed, 22 May 2013 18:03:37 GMT

I'm reading about Ehrenfest's paradox where one considers a rotating disk. If one let r' be the radius of the disc in an inertial frame and r be the radius of the disc when it is at rest. Then the periphery must be Lorentz' contracted such that ##2\pi r' < 2\pi r##, but since the radial line is perpendicular to the direction of motion of the disk it is not Lorent'z contracted so that ##r'=r##. Thus the paradox.

The supposed kinematical solution is to consider the disk being accelerated up to a given angular velocity. By an analysis involving the relativity of simultaneity it is found that it is inconsistent to require that the disk be accelerated up to the angular velocity and additionally require it being done while keeping the periphery 'Born rigid'. I.e. accelerated in such a way that the rest length of each line element along the periphery remains constant.

This is claimed to resolve the paradox and my question is why it does that.
More specifically; why does one need to assume a 'Born rigid' acceleration of the disk in order to accept the paradox?

What keeps the planets in orbit

G.U.T.finder — Wed, 22 May 2013 16:01:49 GMT

If the reason why the planets orbit around the sun is the deviations in the fabric of spacetime,what keeps the planets from crashing into each other? Like if I put a boulder on a trampoline and then I put a baseball next, the baseball would go toward the boulder. I think it could be dark matter pushing the planets away and gravity bringing them closer in equilibrium. But who knows the correct answer ( if there is one).

Father-Son Paradox

kannank@live — Wed, 22 May 2013 14:53:46 GMT

This is an extension of twin paradox of STR. Suppose when the father was 30yrs old, he got into this flight which moves close to c. After a while in that he comes back to see his Son older to him! I'm sure this is a possibility from time dilation. Is this really possible? If so, how is the father going to explain all this to someone else who doesn't know about his space trip!! :-)

Timelike Killing vector field and stationary spacetime

maxverywell — Tue, 21 May 2013 10:53:01 GMT

I am trying to understand why in the definition of a stationary spacetime the Killing vector field has to be timelike.

It is required that the metric is time independent, i.e. the time translations [itex]x^0 \to x^0 + \epsilon[/itex] leave the metric unchanged. So the Killing vector is [itex]\xi^{\mu}=\delta_{0}^{\mu}[/itex]. In other words it is required that there exists a Killing vector that has only [itex]x^{0}[/itex] component and no space components (they are 0). But in general, a timlike vector can also have non zero space components. For example, the vector [itex](1,1/2)[/itex] in two dimensional Minkowski space, is timelike because [itex](1,1/\sqrt{2})^2=-1/2<0[/itex]. This vector is a linear combination of the two independent killing vectors (1,0) (time translations) and (0,1) (space translations) of the 2-d Minkowski spacetime, so it's also a killing vector and it generates translations in time and also in space.

So, if there exists a timelike killing vector with non zero space components, then necessarily there exists the killing vector [itex]\delta_{0}^{\mu}[/itex]. So saying that there should exist timelike Killing vector is because of this? It is simply a more general statement than saying that "spacetime is stationary if it admits the Killing vector field [itex]\delta_{\mu}^{0}[/itex]"?

Geometric description of the free EM field

dextercioby — Tue, 21 May 2013 04:58:38 GMT

I'm not a geometer, so I beg for indulgence on the below:

In a modern geometrical description of electromagnetism (either in flat or in curved space-time*), I see at least 3 (or 4) (fiber) bundles over the 4D space-time taken to be the base space:

* 1 the cotangent bundle and the bundle of p-forms over space time (de Rham complex). Differential operator: d Here the EM field appears as: the potential is a 1 form field, the field strength is a 2 form field. F=dA.

* 2 the (principal?)[itex] SO_{\uparrow}(1,3) [/itex] bundle of tensors over space time. Here if the group is replaced by SL(2,C), we get the spinor bundle over space time, a concept described in the 13th chapter of Wald's book - in the context of a curved space time. Here I imagine A and F as covariant and double covariant tensors. Differential operator = [itex] \partial_{\mu} [/itex] or [itex] \nabla_{\mu} [/itex].

* 3 the so-called U(1) gauge bundle over space time (principal/associated bundle ?) , with F and A having particular descriptions. Here we have another differential operator d-bar, such as F = d-bar A.

Questions:

a) How are 1,2,3 exactly related ?
b) Is d-bar at 3 related to d at 1 ? I suspect yes. How can it be proven ? The operator at 2, how's it related to d-bar at 3 or to d at 1 ?

Thank you!

Good books on both Relativities

JamesOrland — Mon, 20 May 2013 18:48:01 GMT

Hello. I'd like to hear opinions on good books about either/both Relativities. I have some knowledge of the workings of Special Relativity, and naught but the barest knowledge of General Relativity beyond the basic verbal descriptions, so I'd like to know about good books on that.

Thank you in advance for your help!

Motion through spacetime

maxverywell — Mon, 20 May 2013 18:33:13 GMT

Ok, I understand what the motion through space is, but I have difficulty in understanding what the motion through spacetime is. Spacetime is a 4 dimensional manifold with time being one of the 4 dimensions.

When a particle moves through spacetime with some velocity, it moves through space with the conventional 3-velocity, but it also moves through time. What does it mean to move through time and how is the velocity defined for the motion through the dimension of time?

Do we live in 3 or 4 dimensions?

Stricklandjr — Mon, 20 May 2013 16:37:19 GMT

pleese don't critical on my little knowledge of this. a few of us have bull sessions sometimes and my cousin points front side up and says we live in 3 dimencions but the whole universe is 4 dimencions. this other guy always says my cousin is full of bull and einstine just used them for calculations. who is right?

also i'm new and read the rules for this forums. what did greg mean by you can't bring up ether theory?

wrong signs Ricci tensor components RW metric tric

enomanus — Mon, 20 May 2013 10:28:06 GMT

Hi, I am working through GR by myself and decided to derive the Friedmann equations from the RW metric w. ( +,-,-,-) signature. I succeeded except that I get right value but the opposite sign for each of the Ricci tensor components and the Ricci scalar e.g. For R00 I get +3R../R not -3R../R . I calmly ignored the wrong signs and substituted the opposite sign into the Einstein equations because 1. they worked and 2. it matched the values for G00 G11 given in textbooks. Obviously I am curious as to what I did wrong! Any ideas?? Thanks !!