General relativity (gravitation) in time and one spatial dimension

I don't have any idea of general relativity but intend to learn. Is it a good idea to study general relativity in two dimensions (time and single spatial dimension) in the begining to get good idea on the subject? If it is, then please give some references for such a treatment.

Answer

Depends on your reference. One thing that should be said is that 1+1 relativity theory is trivially simple in several ways.

First, the Riemann Curvature tensor $R_{abcd}$ is determined completely by one component, meaning that Einstein's equation reduces to a single PDE. Since this is generally a tensor equation involving a somewhat large number of coupled PDE's, you will likely not get appropriate intuition for the theory.

Second, there is a conformal invariance$^{1}$ built into the 1+1 Einstein's equations that further makes the theory very easy to solve, since we're now dealing with a single PDE (plus a matter equation of state) with conformal invariance that lets us control the input o that PDE.

These two complications will give a very bad intuition for how Einstein's equations work--in particular, they reduce Einstein's equation to an equation depending only on topological invariants, and not upon local properties of the theory. There can be no radiation in 1+1 gravity, for example. In 3+1 dimensions, where neither of the above simplifications apply, the qualitative behaviour of the theory is very different.

That said, a guided study of particular Minkowskian two-spaces can help one understand full Relativity theory, but that is something best done for you by an expert. Please just start with Schutz's book if you're just trying to teach yourself Relativity.

$^{1}$conformal invariance means I can map the metric $g_{ab}$ into $\bar g_{ab} = \phi g_{ab}$ and have $\bar g_{ab}$ satisfy the same equation that $g_{ab}$ did. Einstein's equations in 1+1 dimensions are unable to distinguish between the metric and a rescaling of the metric.

newtonian gravity - Dimensional analysis - When can you introduce constants that make dimensions compatible?

I have just read this question: What justifies dimensional analysis. One statement was:

Maybe the speed of a comet is given by its period multiplied by its mass. Why not?

As a formula this is $v=mT$. How do we know that this is wrong? I am not asking for the standard answer concerning the incompatibility of the dimensions. Suppose $v\propto m$ and $v \propto T$, then one could argue that $v\propto mT$ and so $v=CmT$ where $C$ is a constant that fixes the dimensions.

I will give another example: $F_G=Gm_1 m_2/r^2$ - Newton's law of gravity. As far as I know, Newton knew the following: $F \propto m_1m_2/r^2$ and he didn't know the value of $G$ so he simply stated $F=Gm_1m_2/r^2$ with $G$ fixing the dimensions.

Now comes my real question: My intuition tells me that $v\propto mT$ is not right but can you exclude it with dimensional analysis? Then you would also have to deny Newton's law of gravity. How do you know that $v=CmT$ is incorrect but $F=Gm_1m_2/r^2$ is not? Especially: When can you "invent" a constant in dimensional analysis which fixes the dimensions? If you could do it all the time, dimensional analysis would not be helpful...

general relativity - Would there be time dilation at the point where two gravitational fields cancel each other out?

My question is very simple, and most likely a stupid one:

One observer is at a point in space were the gravitational force form massive bodies (or a single massive body) cancel each-other out. The second observer is in another hypothetical situation where there exists no massive bodies and therefore no gravity. My question is this, is there a relativistic time difference between the two?

Answer

In the "weak field limit" where the graviational forces are small (such as anything in the solar system, and basically anything not right next to a black hole), the time dilation relative to a distant observer is:

$\Delta T/\Delta T_0 = 1 - \Phi/c^2$.

here $\Delta T_0$ is the time elapsed for an observer at infinity, $\Delta T$ is that time elapsed at some point in the system, and $\Phi$ is the gravitational potential at that point. As @leftaroundabout stated correctly the important factor is the gravitational potential not gravitational forces. Since potentials add, instead of cancelling like the forces, we get twice the time dilation with two planets than we get with one, as @Jim Graber said.

homework and exercises - Quadratic drag projectile motion

I have calculated formulas with 1 dimensional trajectory motion (free-fall) including quadratic drag, and have created the following equations.

These equations of motion are not of much use on its own, therefore I would like either a analytical method/numerical method in order to plot 2-dimensional projectile motion on a graph, y against x. However, I am aware that the equations of motion for two-dimensional projectiles are the following: $$m*a_x=-k*\sqrt{v_x^2+v_y^2}*v_x$$ $$m*a_y=-mg-k*\sqrt{v_x^2+v_y^2}*v_y$$ I have realized that due to the formulas having a circular reference, there will be no general solution, hence there is no analytical method to plot this graph, as the drag in the x-direction will slow down the projectile and will change the drag in the y-direction as well as vise-versa.

Hence, this is where I do not know the approach to the problem, I am not very familiar with numerical methods of solving equations. Is there any way to do this in a spreadsheet or in MATLAB, maintaining a great level of accuracy (If possible, using RK4).

Note: x-velocity is oriented on the right hand side, and the y-velocity directly upwards.

Please correct me if I have made any mistakes.

Answer

$\newcommand{\dd}{\mathrm{d}}$ You basically have two ODEs to solve: \begin{align} \frac{\dd v^\mu}{\dd t}&=\frac{1}{m}F(x^\mu,v^\mu) \tag{1} \\ \frac{\dd x^\mu}{\dd t}&=v^\mu\tag{2} \end{align} which is pretty much the case for most forces in Newtonian mechanics. In order to solve this numerically, you want to discretize space & time. With such a system as (1) & (2), we really only need to worry about slicing up time.

One of the more stable routines is not actually RK4, but a variation of the leapfrog integration called velocity verlet. This turns (1) & (2) into a multi-step process: \begin{align} a_1^\mu &= F\left(x^\mu_i\right)/m \\ x_{i+1}^\mu &= x^\mu_i + \left(v_i + \frac{1}{2}\cdot a_1^\mu\cdot\Delta t\right)\cdot\Delta t \\ a_2^\mu & =F\left(x^\mu_{i+1}\right)/m \\ v_{i+1}^\mu &= v^\mu_i + \frac{1}{2}\left(a_1^\mu+a_2^\mu\right)\cdot\Delta t \end{align} which is actually kinda easy to implement numerically, it's literally just calling the function for the force and then updating a couple arrays (x,y,vx,vy).

Where your problem differs is that $a^\mu=a^\mu\left(x^\mu,\,v^\mu\right)$, which makes computing the second acceleration a bit tricky since $a_2$ depends on $v_{i+1}^\mu$ and vice versa. This answer at GameDev (definitely worth the read for some numerics aspect to the problem) suggests that you can use the following algorithm

\begin{align} a_1^\mu &= F\left(x^\mu_i,\,v_i^\mu\right)/m \\ x_{i+1}^\mu &= x^\mu_i + \left(v_i + \frac{1}{2}\cdot a_1^\mu\cdot\Delta t\right)\cdot\Delta t \\ v_{i+1}^\mu &= v_i^\mu + \frac{1}{2}\cdot a_1^\mu\cdot\Delta t \\ a_2^\mu & =F\left(x^\mu_{i+1},\,v_{i+1}^\mu\right)/m \\ v_{i+1}^\mu &= v^\mu_i + \frac{1}{2}\cdot\left(a_2^\mu-a_1^\mu\right)\cdot\Delta t \end{align} though the author of that post states,

It's not quite as accurate as fourth-order Runge-Kutta (as one would expect from a second-order method), but it's much better than Euler or naïve velocity Verlet without the intermediate velocity estimate, and it still retains the symplectic property of normal velocity Verlet for conservative, non-velocity-dependent forces.

Since this is projectile motion, $x=y=0$ is probably a natural choice for initial conditions, with $v_y=v_0\sin(\theta)$ and $v_x=v_0\cos(\theta)$ as is normal.

If you have a single atom in a box, what state of matter would it be in?

If the walls of the box were made of a completely inert material, what state of matter would the atom be in? Or would it not have one? I realized this is a duplicate of several other questions.

quantum field theory - Regarding a small step in the derivation of the LSZ formula

I'd like to prove the LSZ formula, but there is a specific step that is bugging me a lot. I know there are many subtleties in its derivation, but I'm not worrying about this right now: I'm trying to understand the naive proof, so to speak.

You can find an example of the usual proof here: http://isites.harvard.edu/fs/docs/icb.topic473482.files/06-LSZ.pdf

My question is: how to get eqs. (24-25) from eq. (22-23). In (22-23) the time-ordering symbol is to the left of $(\partial^2+m^2)\phi$ and in (24-25) it is to the right of the differential operator.

I'm asking how to go from $T(\partial^2+m^2)\phi\cdots$ to $(\partial^2+m^2) T\phi\cdots$. I feel that this cannot be done in general, because the symbol $T$ will introduce some $\Theta(x_0-y_0)$ functions, which when differentiated will give rise to some deltas.

This step makes no sense to me... Is my question legitimate? Is there anything from the proof I'm missing?

If I think of specific examples, I find different results depending on whether the $T$ symbol is to the right or to the left of the Klein-Gordon differential operator. This means that the order is important... so, which is the right order? Should the $T$ symbol be placed to the right or to the left of the KG operator? In the begining of the proof the $T$ symbol is always to the left, and in the end, it is always to the right.

This same problem appears on many proofs online, such as

http://isites.harvard.edu/fs/docs/icb.topic473482.files/06-LSZ.pdf (eqs. 22-25)

http://www2.ph.ed.ac.uk/~egardi/MQFT_2013/MQFT_2013_lecture_2.pdf (eqs. 32-33)

Srednicki's book, page 51 (eqs. 5.14-5.15) online: http://web.physics.ucsb.edu/~mark/qft.html

etc.

Edit: I was asked to post a self-contained question, so I'll write the delails here:

Take the in state to be $|i\rangle\propto a^\dagger_1(-\infty)a^\dagger_2(-\infty)|0\rangle$ and the out state to be $|f\rangle\propto a^\dagger_3(+\infty)a^\dagger_4(+\infty)|0\rangle$. Then the transition amplitude is $$ \langle i|f\rangle \propto \langle 0|a_2(-\infty)a_1(-\infty)a^\dagger_3(+\infty)a^\dagger_4(+\infty)|0\rangle= $$ $$ =\langle 0|Ta_2(-\infty)a_1(-\infty)a^\dagger_3(+\infty)a^\dagger_4(+\infty)|0\rangle $$

Now write $a(+\infty)-a(-\infty)\propto \int \mathrm d x\ \mathrm e^{\cdots}(\partial^2+m^2)\phi$; all terms with $a$'s or $a^\dagger$'s annihilate the vacuum, so that the only remaining term is $$ \langle i|f\rangle \propto \langle 0|T \int \mathrm dx_1\mathrm dx_2\mathrm dx_3 \mathrm dx_4\ \mathrm e^{\cdots}(\partial_1^2+m^2)(\partial_2^2+m^2)(\partial_3^2+m^2)(\partial_4^2+m^2) \phi_1\phi_2\phi_3\phi_4 |0\rangle $$

This is usually writen as $$ \langle i|f\rangle \propto \int \mathrm dx_1\mathrm dx_2\mathrm dx_3 \mathrm dx_4\ \mathrm e^{\cdots}(\partial_1^2+m^2)(\partial_2^2+m^2)(\partial_3^2+m^2)(\partial_4^2+m^2) \langle 0|T\phi_1\phi_2\phi_3\phi_4 |0\rangle $$ but this step is problematic, because the $T$ symbol and the KG operator don't commute, right?

Answer

Comments to the question (v4):

1. 
   

   OP is wondering about the contact terms from commuting time-derivatives and time-ordering symbol $T$, cf. e.g. this and this Phys.SE posts.

   
   


2. 
   
   

   Consider the on-shell $S$-matrix side of the LSZ reduction formula. The time-differentiation from the boundary terms $$ T\left[\prod_{i=1}^n \left\{ a_{{\bf p}_i}^{\#}(t_i\!=\!\infty) -a_{{\bf p}_i}^{\#}(t_i\!=\!-\infty)\right\}\right] $$ $$~=~T\left[\prod_{i=1}^n \int_{\mathbb{R}}\! \mathrm{d}t_i~\frac{d}{dt_i}a_{{\bf p}_i}^{\#}(t_i)\right] ~=~\left[\prod_{i=1}^n \int_{\mathbb{R}}\! \mathrm{d}t_i~\frac{d}{dt_i}\right]T\left[\prod_{j=1}^n a_{{\bf p}_j}^{\#}(t_j)\right]\tag{A}$$ can be moved outside the time-ordering symbol $T$ because the contact terms vanish $$\delta(t_i-t_j) \left[a_{{\bf p}_i}^{\#}(t_i) ,a_{{\bf p}_j}^{\#}(t_j) \right]~=~0\tag{B} $$ for generic$^1$ 3-momenta ${\bf p}_i\neq {\bf p}_j$. Eq. (B) follows from locality, i.e. spatially separated operators commute. [Here $\#$ refers to creation/annihilation operators, i.e. with or without Hermitian conjugate.]

   
   


3. 
   

   In the Hamiltonian formulation with only first-order time-derivatives, the above shows that contact terms vanish.

   
   


4. 
   

   In the Lagrangian formulation with second-order time-derivatives (i.e. one more time differentiation, which is the case OP is asking about), one may show using similar arguments, that contact terms do not contribute to the S-matrix. See also e.g. Ref. 1.

   
   

References:

1. M.D. Schwartz, QFT and the Standard Model; Section 6.1, p.72, footnote 2.

--

$^1$ Also note that we are usually not interested in disconnected parts of the $S$-matrix, which implies more momentum conservation laws, and hence more special values of the momenta, such as, e.g., ${\bf p}_i={\bf p}_j$.

The Matter-Vacuum Boundary in General Relativity

A previous Stack question (before I joined) asking about continuity in GR received replies which suggested that Curvature would be discontinuous at say a planetary boundary (assume no atmosphere for simplicity). I will analyse some basics of this and then return to that question.

It is true that the Stress-Energy Tensor $T{_a}{_b}=0$ outside the body and is nonzero in the interior resulting in a discontinuity at the surface. This would imply that the Ricci Tensor $R{_a}{_b}$ is also discontinous at the boundary, and zero in the vacuum part as expected from the Einstein equations. However the Riemann Curvature Tensor $R{_a}{_b}{_c}{_d}$ (which generates the physically measurable accelerations) has contributions from the Weyl Curvature Tensor $C{_a}{_b}{_c}{_d}$ as well. In fact the Ricci Tensor "hands over" to the Weyl Tensor at the boundary: thus the Riemann Tensor stays non-zero there. However this "hand over" does not imply continuity, unless there is some GR Theorem which says that the Riemann Tensor stays continuous in this region.

Also in the Newtonian approximation the analogous role is played by the gravitational potential $\phi$ in the Poisson equation $\nabla^2 \phi = 4 \pi G\rho$. Clearly this shows a discontinuity too as the density $\rho$ suddenly drops off at the boundary. However the discontinuity is in the second derivative of the potential: the potential itself is continuous. This means that in exiting a planetary cave or mine one does not suddenly meet a change in Gravitational potential.

However I do not know any theorem in GR which guarantees such continuity. The applicable in-the-large scenario might be the surface of a neutron star; there may be in-the-small particle models too.

Answer

Roy, your wishful thinking is manifestly impossible. If the tensor $T_{\mu\nu}$ is discontinuous, and it surely is on the surface of a solid, then Einstein's equations guarantee that the Einstein tensor $G_{ab}$ is discontinuous as well - up to a normalization, it's the same tensor. It follows that the Ricci tensor and Riemann tensor, $R_{\mu\nu}$ and $R_{\kappa\lambda\mu\nu}$, must also be discontinuous because the Einstein tensor $G_{\mu\nu}$ can be easily calculated both from the Ricci tensor as well as from the Riemann tensor, so if the Ricci or Riemann tensor were continuous, the Einstein tensor would have to be continuous, which is an obvious contradiction.

I just proved the opposite theorem that the Riemann tensor is discontinuous.

You should realize that the Riemann tensor has a higher number of components than the Ricci (or Einstein) tensor, so its continuity - which means the continuity of all of its components - is an even stronger condition than the continuity of the Ricci (or Einstein) tensor. The argument above proves that none of these tensors is continuous in the presence of solids - which is why there can't be any theorem saying the opposite thing (it would be wrong). Another question is whether the Weyl tensor is continuous near such boundaries. I don't know the answer. The answer could be easily calculated from the very formula for the Weyl tensor.