# An argument that massive particles don't redshift?

1. Nov 20, 2015

### jcap

$$ds^2=a^2(\eta)(d\eta^2-dx^2-dy^2-dz^2)$$
This metric has the following non-zero Christoffel symbols:
\begin{eqnarray*}
\Gamma^0_{\alpha \beta} &=& \frac{\dot{a}}{a} \delta_{\alpha \beta} \\
\Gamma^i_{0j} &=& \Gamma^i_{j0} = \frac{\dot{a}}{a} \delta^i_j
\end{eqnarray*}
where $\alpha,\beta=\{0,1,2,3\}$ and $i,j=\{1,2,3\}$.

Let us assume the following:

1. A massive particle travels on a geodesic $x^\mu(\lambda)$ with tangent vector $U^\mu=dx^\mu/d\lambda$ where $U^\mu U_\mu=1$.
2. The 4-momentum of the particle is given by $P^\mu = m U^\mu$.
3. A timelike vector field (not assumed to be a Killing field) $\eta^\mu=\delta^\mu_0$, $\eta_\mu=\eta^\nu g_{\mu\nu}=\delta^\nu_0g_{\mu\nu}=g_{\mu0}$.

The energy of the particle, $E$, with respect to the co-ordinate system is given by:
$$E=\eta_\mu P^\mu=m\eta_\mu U^\mu$$
As the vector $\eta^\mu$ is not assumed to be a Killing vector then the energy $E$ is not necessarily constant.

Now the rate of change of the energy $E$ along the worldline $x^\mu(\lambda)$ is given by:
\begin{eqnarray*}
U^\nu \nabla_\nu(m\eta_\mu U^\mu)&=& mU^\nu U^\mu \nabla_\nu \eta_\mu + m\eta_\mu U^\nu \nabla_\nu U^\mu \\
&=& mU^\nu U^\mu \nabla_\nu \eta_\mu
\end{eqnarray*}
where the last term on the right-hand side is zero as the particle moves on a geodesic.

We now evaluate the covariant derivative in the above equation using $\eta_\mu=g_{\mu0}$:
\begin{eqnarray*}
\nabla_\nu \eta_\mu &=& \partial_\nu \eta_\mu - \Gamma^\lambda_{\nu \mu} \eta_\lambda \\
&=& \partial_\nu g_{\mu0} - \Gamma^0_{\nu\mu} g_{00} \\
&=& \partial_\nu g_{\mu0} - \frac{\dot{a}}{a} \delta_{\nu\mu} g_{00} \\
&=& \mbox{diag}(a \dot{a},-a\dot{a},-a\dot{a},-a\dot{a}) \\
&=& \frac{\dot{a}}{a}g_{\nu\mu}
\end{eqnarray*}
Thus the rate of change of the energy of the particle along the geodesic is given by:
\begin{eqnarray*}
U^\nu \nabla_\nu(m\eta_\mu U^\mu) &=& m\frac{\dot{a}}{a}U^\nu U^\mu g_{\nu\mu}\\
&=& m\frac{\dot{a}}{a} \\
&=& m \frac{(da/d\tau)(d\tau/d\eta)}{a} \\
&=& m \frac{da}{d\tau}
\end{eqnarray*}
If I integrate both sides of the above equation I find that the energy $E$, with respect to the metric co-ordinates, of any massive particle in geodesic motion is given by:
\begin{eqnarray*}
E = m\eta_\mu U^\mu &=& \eta_\mu P^\mu \\
&=& \eta^\mu P_\mu \\
&=& \delta^\mu_0 P_\mu \\
&=& P_0 \\
&=& ma
\end{eqnarray*}
Now let us calculate the energy $E_{obs}$ that a co-moving observer with 4-velocity $U^\mu_{obs}=\frac{1}{a}\delta^\mu_0$ measures when he observes the massive particle.
\begin{eqnarray*}
E_{obs} &=& U^\mu_{obs} P_\mu \\
&=& \frac{1}{a}\delta^\mu_0 P_\mu \\
&=& \frac{1}{a} P_0 \\
&=& \frac{1}{a} m a \\
&=& m
\end{eqnarray*}
Notice that the energy $E_{obs}$, that a co-moving observer measures, is constant for any massive particle moving along a geodesic and not just constant for a particle that is at rest in the co-moving frame.

Therefore it seems that massive particles do not redshift as they travel on geodesics through the expanding Universe which is contrary to received wisdom.

Where have I gone wrong?

2. Nov 20, 2015

### Chalnoth

I'm not entirely sure. But it's definitely incorrect.

You should be able to get the correct answer using only classical arguments, showing that if I throw a ball at some velocity v from the origin in a uniform, expanding universe, that ball will remain at a constant velocity with respect to me, but will get slower and slower with respect to the comoving observers it passes along its journey.

3. Nov 20, 2015

### Staff: Mentor

No, it isn't. The vector $\eta_\mu$ is not a unit vector. You need to have a unit timelike vector for this to work. The unit timelike vector in the direction of $\eta_\mu$ will be the 4-velocity of a comoving observer, $v^{\mu} = (1 / a) \eta^{\mu}$. The corresponding 1-form will be $v_{\mu} = g_{\mu \nu} v^\nu = (1 / a) g_{\mu 0} = (1 / a) \eta_\mu$. The rate of change of $E$ along the geodesic worldline of the particle is then $U^\nu \nabla_\nu \left( m v_{\mu} P^{\mu} \right) = U^\nu \nabla_\nu \left( m (1 / a) \eta_{\mu} P^{\mu} \right)$. This should lead to a predicted "redshift".

4. Nov 21, 2015

### jcap

I agree that I need a unit timelike vector if the particle energy is measured by a local co-moving observer. But I'm defining $E$ as the energy of the particle with respect to the co-ordinate system itself or in other words with respect to a "reference" observer at the present time $\eta_0$.

Maybe such an idea doesn't make sense.

5. Nov 21, 2015

### jcap

But GR says that the cosmological redshift of photons for example is really due to time dilation which can't be explained by Newtonian physics.

6. Nov 21, 2015

### marcus

Are you certain? Do you have a link to a source where some GR expert says this? It seems like a strange idea, and one I don't remember ever seeing put forward before.

Why wouldn't the cosmological redshift be the simple consequence of the metric expansion of distances (e.g. applied to wave propagation according to Maxwell's equation) during the time the light is in transit? It seems like a no-brainer given that distances and wavelengths are enlarged by exactly the same factor, during transit:

arec/aem = z +1

Between the time the light is emitted and then received by us, cosmological distances grow by the LHS factor.
And wavelengths are enlarged by the RHS factor, z+1.

Last edited: Nov 21, 2015
7. Nov 21, 2015

### Staff: Mentor

Energy with respect to any observer, "reference" or not, is the contraction of the object's 4-momentum with the observer's 4-velocity, which is always a unit vector. (In the case of a "reference" observer that is comoving, that 4-velocity is the one I wrote down.) There is no such thing as "energy with respect to the coordinate system" that does not require a unit vector.

8. Nov 21, 2015

### jcap

A "reference" observer at the present conformal time $\eta_0$ when $a(\eta_0)=1$ will have a unit 4-velocity $v^\mu=\frac{1}{a(\eta_0)} \eta^\mu=\eta^\mu$.

9. Nov 21, 2015

### jcap

Well I'm not sure what GR experts say but the standard derivation of the cosmological redshift involves comparing a time interval at the emission time with a time interval at the receiving time. The wavelengths of the emitted and received light are calculated directly from these time intervals. The expansion of space itself does not enter into the calculation.

For example:

https://en.wikipedia.org/wiki/Redshift#Mathematical_derivation

In terms of conformal time one can say:

$$\Delta \eta_{rec} = \Delta \eta_{em}$$
$$\frac{\Delta t_{rec}}{a(t_{rec})}=\frac{\Delta t_{em}}{a(t_{em})}$$
$$\frac{\lambda_{rec}}{\lambda_{em}}=\frac{c\Delta t_{rec}}{c\Delta t_{em}}=\frac{a_{rec}}{a_{em}}$$

Last edited: Nov 21, 2015
10. Nov 21, 2015

### marcus

==quote==
Describing the cosmological expansion origin of redshift, cosmologist Edward Robert Harrison said, "Light leaves a galaxy, which is stationary in its local region of space, and is eventually received by observers who are stationary in their own local region of space. Between the galaxy and the observer, light travels through vast regions of expanding space. As a result, all wavelengths of the light are stretched by the expansion of space. It is as simple as that..."[38]Steven Weinberg clarified, "The increase of wavelength from emission to absorption of light does not depend on the rate of change of a(t) [here a(t) is the Robertson-Walker scale factor] at the times of emission or absorption, but on the increase of a(t) in the whole period from emission to absorption."[39]
==endquote==
As Weinberg indicates it does not depend on the SPEED of the galaxy. It is not a special relativistic time dilation effect.
The distance to a typical galaxy that we can see with telescope is increasing FASTER THAN C. So it would be impossible to calculate that kind of time dilation.
Think of the source galaxy, and our Milkyway galaxy, as not moving in their surrounding spaces, and the distance between them simply increasing (metric expansion). Nobody gets anywhere by the pattern of distance expansion, no goal is approached, it is not like ordinary relative motion that is covered in SR. In SR space does not expand.

11. Nov 21, 2015

### jcap

I'm saying that cosmological redshift is due to gravitational time dilation not to special relativistic time dilation. One can see this most easily when the FRW metric is expressed in conformal co-ordinates with $g_{00}=a^2$ giving an increasing gravitational potential. Cosmological redshift is analogous to the gravitational redshift of light coming out of a potential well.

Last edited: Nov 21, 2015
12. Nov 21, 2015

### Staff: Mentor

But this is just one point on the reference observer's worldline. You're trying to calculate the rate of change of energy of an object with respect to comoving observers; that means you need to use their 4-velocity all along their worldlines, not just at one point.

To put it another way, the object will cross the worldline of one reference observer at $\eta_0$, but it will cross the worldline of another reference observer at a different time $\eta_1 \neq \eta_0$. So you can't calculate the rate of change of the object's energy by using the 4-velocity at $\eta_0$ alone; you need the 4-velocity at $\eta_1$ as well, and that is not $\eta^\mu$. Your calculation effectively assumes that the object crosses every comoving observer's worldline at the same conformal time, which is impossible.

13. Nov 21, 2015

### Staff: Mentor

This is not correct. The concept of gravitational time dilation only makes sense in a stationary spacetime. FRW spacetime is not stationary.

The metric coefficient $g_{00}$ can only be interpreted as "gravitational potential" in a stationary spacetime. FRW spacetime is not stationary.

Incorrect. See above.

14. Nov 22, 2015

### Chalnoth

Certainly you can't get the right answer with relativistic particles, but the question in the OP was about massive particles. Running through the calcualtions classically, it should be quite easy to come up with an equation that is correct for particles whose motion compared to the comoving coordinates is non-relativistic. This, to me, would serve as a good check to make sure I'd done my math right in the relativistic case.

15. Nov 25, 2015

### nikkkom

I did not look at your math, but for massive particles, "redshift" is simply the decrease of their velocity relative to comoving coordinates.

As it moves, any massive particle with nonzero velocity in the expanding Universe reaches locations where its velocity is smaller and smaller. In the limit of infinite future, it will reach a place where its comoving velocity falls to zero.