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Why does mass warp spacetime? 
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#19
Jan2013, 09:03 AM

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Although there is no real scientific answer to your question, string theory offers some interesting insights. [That mass and its equivalent, gravity, can affect space and as well the relative passage of time is one of most profound findings of all time. It's downright 'crazy' based on our everyday intuitions.]
In string theory, fundamental components of particles, strings, are vibrating energy modes. It turns out that these interact with the degrees of freedom, or geometrical dimensions, in which we all find ourselves. Different sizes and shapes of additional dimensions can be mathematically associated with different characteristics of strings: varying vibration patterns correspond to things like particle size, charge, spin that we observe macroscopically. So strings and geometry, spacetime, interact analogous to mass/energy in general relativity. This offers some insights, perhaps, why not only mass, but also energy and momentum density warps spacetime. 


#20
Jan2013, 09:20 AM

P: 5,632

I find Greene's above description along the lines of the 'rubber sheet' analogy for gravity...[which Greene discuss right after in his book] or the 'balloon analogy' for cosmology, useful as a perspective, but they all come with limitations. 


#21
Jan2013, 10:48 AM

P: 642

Though this has nothing, or almost nothing, to do with gravity. 


#22
Jan2013, 12:00 PM

Emeritus
Sci Advisor
PF Gold
P: 9,248

In my own rest frame, my world line coincides with the time axis. So the tangent to my world line is in the 0 direction (axes numbered from 0 to 3, with "time" being 0). Every vector of the form $$\begin{pmatrix}r\\ 0\\ 0\\ 0\end{pmatrix}$$ where r is a real number is a tangent vector to the world line. The tangent vector with Minkowski "norm" 1 (c for those who don't set c=1) is called my fourvelocity. I will denote its coordinate matrix in my own rest frame by u. We have $$u=\begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix},\qquad u^2=u^T\eta u=\begin{pmatrix}1 & 0 & 0 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix}=1.$$ Greene calls ##\sqrt{u^2}## the speed through spacetime. We have ##\sqrt{u^2}=1##. If we restore factors of c, this is ##\sqrt{u^2}=c##. (Greene uses the metric signature + instead of ++++, so when he does this, he gets ##u^2=c^2##, and is therefore able to write the speed through spacetime as ##\sqrt{u^2}##. His ##u^2## is equal to my ##u^2##). Let's boost my four velocity to the rest frame of an observer who has velocity v in my coordinate system. I should have velocity v in his. $$u'=\Lambda(v)u=\gamma\begin{pmatrix}1 & v^1 & v^2 & v^3\\ v^1 & * & * & *\\ v^2 & * & * & *\\ v^3 & * & * & *\end{pmatrix}\begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix}=\gamma\begin{pmatrix}1\\ v^1\\ v^2\\ v^3\end{pmatrix}.$$ The asterisks denote matrix elements that are irrelevant to what we're doing here. If anyone cares, they are the components of the 3×3 matrix $$\frac{1}{\gamma}I+\left(1\frac 1 \gamma\right)\frac{vv^T}{v^Tv}.$$ The velocity components can be calculated like this: $$\frac{dx^i}{dt^i}=\frac{u'^i}{u'^0}=\frac{\gamma v^i}{\gamma}=v^i.$$ As expected, my velocity in the new coordinate system is minus the velocity of the boost. This result is the reason why the normalized tangent vector is called the fourvelocity. The world line is the range of a curve ##C:\mathbb R\to M## where M is Minkowski spacetime. Its representation in a global coordinate system ##x:M\to\mathbb R^4## is the curve ##x\circ C:\mathbb R\to\mathbb R^4##. The world line is said to be parametrized by proper time if the curve C that we use to represent it has the property that for each point p on the world line, the number ##\tau## such that ##C(\tau)=p##, is the proper time along the curve from C(0) to p. Such a C has the advantage that the fourvector with components ##(x\circ C)^\mu{}'(t)## is automatically normalized. So if y is my rest frame, and x is the coordinate system we transformed to above, we have ##u^\mu=(y\circ C)^\mu{}'(\tau)## and ##u'^\mu=(x\circ C)^\mu{}'(\tau)##. It's conventional to denote ##(x\circ C)^\mu(\tau)## by ##dx^\mu/d\tau##, so we have $$u'^\mu=\frac{dx^\mu}{d\tau}.$$ Now let's use the fact that ##u^2## is Lorentz invariant. $$1=u^2=u'^2 =(u^0)^2+(u^1)^2+(u^2)^2+(u^3)^2 =\left(\frac{dt}{d\tau}\right)^2+\sum_{i=1}^3 \left(\frac{dx^i}{d\tau}\right)^2.$$ Let's manipulate this result with some nonrigorous physicist mathematics. (These things can of course be made rigorous). \begin{align} &\frac{dt}{d\tau} =\sqrt{1+\sum_{i=1}^3\left(\frac{dx^i}{d\tau} \right)^2}\\ &\frac{d\tau}{dt} =\frac{1}{\sqrt{1 +\sum_{i=1}^3\left(\frac{dx^i}{d\tau} \right)^2}}\\ &1=\left(\frac{d\tau}{dt}\right)^2 \left(1+\sum_{i=1}^3\left(\frac{dx^i}{d\tau} \right)^2\right) =\left(\frac{d\tau}{dt}\right)^2+\sum_{i=1}^3 \left(\frac{dx^i}{dt}\right)^2. \end{align} Greene calls the square root of the first term the speed through time and the square root of the second term the speed through space. (This is according to note 6 for chapter 2 (p. 392) of "The elegant universe"). This allows him to say that an increase of the speed through space must be accompanied by an decrease of the speed through space. If we had been talking about the motion of a massless particle (i.e. light) instead of the motion of an observer, we would have had ##u^2=0## instead of ##u^2=1##. It's easy to see that what this does to the calculation above is to eliminate the first term on the righthand side above. Since ##\tau=0## along the world line of a massless particle, this means that the result $$\left(\frac{d\tau}{dt}\right)^2+\sum_{i=1}^3 \left(\frac{dx^i}{dt}\right)^2=1$$ holds for massless particles too. 


#23
Jan2013, 01:02 PM

P: 5,632

Fredrick:
it is A way to view the Lorentz transforms.....that space and time 'morph' into each other as a result of speed...are seen differently by different observers....in his explanation is the [unstated, I think] assumption that space and time remain a fixed background. He also uses it to explain that acceleration is a curve in spacetime, fixed velocity plots as a straight line, while rotational motion appears as a corkscrew. It is easy to picture yourself riding along in such situations where you 'see' spacetime different from your neighbor, and they different from you. He does NOT [and cannot] use it to explain the dynamical nature of spacetime due to mass,energy or gravity. 


#24
Jan2113, 08:36 AM

P: 184

I haven't thought about this for too long, so it might not be an air tight argument, but hopefully it's a good qualitative explanation.
Since you have the equivalence principle, you cannot distinguish acceleration from gravity (effect of spacetime) directly. However, you can distinguish indirectly, by observing matter around you, which is a subject of common confusion for students first learning GR (if you had a window in your accelerating elevator, you could see things accelerating outside of it which would be still if it is only gravitation). So the distribution of matter is important for probing the structure of spacetime. This suggests that one can write a relation between the stressenergy tensor (a tensor that contains the information about the matter distribution of the universe) and the metric (another tensor that contains the information about the structure of spacetime). Note, we have not yet implied that matter warps spacetime, only that since you can use matter to measure spacetime, the two are related somehow, and we simply propose that there is a relation. From here we follow with some mathematics, and one thing we know about matter, which is that 4momentum is conserved. In the language of the stressenergy tensor, this means that the divergence of the stressenergy tenser is 0. So we write one side of the relation as the stressenergy tensor, and the other side we write the most general mathematically sensible tensor we can make out of the metric, but also make sure that this general tensor has the property that it's divergence is zero. If you do this you get einstein's equation, at which point you have GR and gravity warping spacetime etc. So it's all an observation that the equivalence principle makes it so that the only way you can measure spacetime is through matter distributions. Also, the idea that matter warps spacetime is a little misleading. The stress energy tensor typically depends on the metric, and thus depends on the spacetime itself. So spacetime and matter are determined simultaneously. It's better to think that two are closely related, but one does not cause the other. Quantum mechanically this might change though (for example, string theory is typically written as perturbations of a fixed background spacetime), but classically this is the most "complete" understanding of GR. 


#25
Jan2213, 05:08 PM

P: 308

hmmm  doesnt the "fact" that higgs bosons have been "discovered" now, which disconnects mass as an instrinsic aspect of matter, mean that there is some particle nature of gravity, or some connection between the action of the higgs boson and the gravitational field, or some other ridiculously confusing "explanation" for gravity now? i am unable to fathom how gravity is merely a warped field since we have now introduced the idea that the higgs boson is responsible for "mass" in some manner that is separate from the matter itself...



#26
Jan2213, 05:32 PM

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#27
Jan2213, 05:45 PM

P: 3,914

dtau^{2} = dt^{2}  dx^{2} to: dt^{2} = dtau^{2} + dx^{2} So instead of a pseudo Euclidean line element dtau, you have an ordinary Euclidean line element dt and tau as a dimension.  proper time rate is a projection of c on the time dimension  spatial speed is a projection c on the space dimension It was used by Lewis C. Epstein in this book Relativity Visualized Here some visualizations based on this: http://www.adamtoons.de/physics/relativity.swf (relation of speed to length contraction and time dilation) http://www.adamtoons.de/physics/twins.swf (visual comparison to the Minkowski interpretation) http://www.physics.ucla.edu/demoweb/...spacetime.html (some of Epstein's original illustrations) http://www.adamtoons.de/physics/gravitation.swf (relation of gravity and gravitational time dialtion) http://www.relativitet.se/Webtheses/lic.pdf (technical discussion in Chapter 6) The neat thing that if you have the direction in spacepropertime, then you can replace: space > momentum propertime > restmass coordinate time > Energy (relativistic mass) E^{2} = m^{2} + p^{2} Which again allows a geometrical interpretation:  rest mass is a projection of energy on the time dimension  momentum is a projection of energy on the space dimension 


#28
Jan2213, 07:44 PM

Mentor
P: 16,975




#29
Jan2313, 04:00 AM

P: 184

So you can claim that ANY particle is related to the gravitational field, so the higgs doesn't hold an especially important place in it. 


#30
Jan2313, 10:57 AM

P: 308

inre: "photons themselves cause gravitational curvature"
i do not think this is true. a photon has no location, and does not cause curvature of spacetime. 


#31
Jan2313, 11:15 AM

Emeritus
Sci Advisor
PF Gold
P: 9,248

Things get more complicated with photons. The term is defined by quantum electrodynamics, so now we're talking about a quantum field's contribution to the metric. We seem to need a quantum theory of gravity to determine that, but there's no reason to think that it would be zero. 


#32
Jan2313, 11:30 AM

P: 184




#33
Jan2313, 02:00 PM

P: 5,632




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