Why does mass warp spacetime?

DaleSpam

This is why the rules of the forum specify mainstream scientific references, not pop-sci references. Do a search for "Brian Greene" on this forum and you will get a taste of the headache that pop-sci treatments cause.

Naty1

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.

Naty1

I've got Brian Greene's book FABRIC OF THE COSMOS where he explains acceleration using the above concepts. I found it useful as ONE perspective, but it seems unpopular here among some. I found it helpful when approaching light cones for the first time....I don't see much difference...

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.

Whovian

Definitely. Depending on what this 4-velocity actually "means," if we manage to interpret it correctly, we're going to be able to predict Special Relativity.

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

Fredrik

Brian Greene actually uses language like that, but (as you can see in his books) it doesn't have anything to do with "warps" (i.e. gravity). He uses it to explain time dilation and other SR phenomena. Here's a quote from "The fabric of the cosmos".

I think this way of looking at it was made popular by Brian Greene. He uses it in both "The elegant universe" and "The fabric of the cosmos". The technical explanation (in units such that c=1, and with a -++++ signature) is as follows.

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 four-velocity. 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 four-velocity.

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 four-vector 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 non-rigorous 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 right-hand 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.

Naty1

Fredrick:

yes,
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 space-time, 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' space-time different from your neighbor, and they different from you.

He does NOT [and cannot] use it to explain the dynamical nature of space-time due to mass,energy or gravity.

DimReg

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 stress-energy 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 4-momentum is conserved. In the language of the stress-energy tensor, this means that the divergence of the stress-energy tenser is 0. So we write one side of the relation as the stress-energy 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.

jnorman

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...

Naty1

Most of the particle mass does NOT come from the Higgs field, but from intrinsic particle energy. Anyway, the QM description [Higgs} is a distinctly different mathematical formalism from GR...we'll have to await 'quantum gravity' to combine them.

A.T.

The interpretation is valid, but apparently not very useful to do math. However it allows some nice, geometrically intuitive explanations. Basically you just rearrange the formula for time-like worldliness:

dtau² = dt² - dx²

to:

dt² = dtau² + dx²

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 space-propertime, then you can replace:
space -> momentum
propertime -> restmass
coordinate time -> Energy (relativistic mass)

E² = m² + p²

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

DaleSpam

I would be careful with this. The term "dimension" doesn't seem to apply to dtau. It isn't a basis vector in any vector space, like dt and dx are.

DimReg

No, the higgs boson doesn't imply anything of that sort. First off, "mass" is not required for gravitation, photons themselves cause gravitational curvature but are massless. Anything with a non-zero stress-energy tensor causes gravitation. The higgs boson does explain features of mass, which helps dictate which particle can decay into which other particle etc. but it doesn't cause gravitation (the same particles without the higgs would still cause gravity).

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.

jnorman

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.

Fredrik

A classical electromagnetic field certainly contributes to the stress-energy tensor of the fields in spacetime, and therefore to the metric (which determines the curvature).

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.

DimReg

Actually, you have it the wrong way around. General relativity works worse for point particles (not test particles) than for spread out fields. By photons, I was refering more to classical light than to the quantum description of the photon. But classically at least, the electromagnetic field, and thus light, has a stress-energy tensor and does in fact cause gravity. The electromagnetic field is the limit of the quantum field the photon is part of, and that's why I say that "photons cause gravitational curvature".

Naty1

Hopefully that result will be the theory that 'corrects' GR and current quantum mechanics near space-time singularities...divergences.....like the big bang and the 'center' of a black hole.