# Non-rest mass, is it real?

1. Jul 17, 2007

### duordi

Suppose there is a mass traveling away from me at 99.99% of the speed of light.

The value of the mass of the object from my perspective can be calculated considering all forms of energy the mass has and also taking into account the rest mass.

Now a light signal is set to me.

The light signal will be red shifted due to a receding velocity of the source.

The light signal will be red shifted additionally due to the rest mass - gravitational field of the object sending the light signal.

Will the light signal be red shifted further by the additional gravitational field due to the receding masses energy of motion as I perceive it to be?

In other words will the “non-rest mass” of an object which is due to a differential velocity between my self and an object cause an additional red shift in a light signal sent to me from the object.

2. Jul 17, 2007

### Schrodinger's Dog

Relativity has shown that objects cause red shift due to both their relative motion Special Relativity(SR) length contraction and time dilation, and a shift due to the non rest mass in General Relativity(GR) which becomes increasingly significant as you approach light speed. So in order to correctly establish the red shift then GR and SR must be taken into account.

http://en.wikipedia.org/wiki/Gravitational_redshift

Taking this into account, then any relative velocity will have an additional shift due to general relativity also, although it may be incredibly small at non relativistic speeds.

Last edited: Jul 17, 2007
3. Jul 17, 2007

### duordi

So red shift can be caused by length contraction?

In that case is it possible that the red shift caused by “receding” distant galaxies could be caused instead, by shortening of our measuring stick because we a shrinking as we spiral into the center of our galaxy and experiencing stronger gravitational fields?

I am assuming that if our measuring stick is getting shorter then the distance to remote objects would seem to be getting larger and that an increasing length would be equivelent to a velocity.

4. Jul 17, 2007

### smallphi

The effect can be easily calculated if you take the body that emits the light signal to be a spherical star or planet. The usual formula for red shift in that situation assume the observer is stationary at constant distance.

The observed light frequency will change when the observer starts moving relative to the star so the relative motion of star and observer definitely has an effect on the observed frequency.

5. Jul 17, 2007

### K.J.Healey

But his questions is, as the earth moves toward the center of the galaxy (does it? I don't know myself, or by what rate), hwo does that change in our spacial curvature affect our local measuing rods, the ones we use to measure the incoming light.

The only guess I'd have is that the increase in energy that comes from the light now having to travel deeper into our galaxies well to reach the earth would cancel out the effect of the meter-stick's length change due to the increased gravity.

6. Jul 17, 2007

### smallphi

Meter sticks do not change in GR. Change with respect to what? The gravitational redshift is due to different time between two crests of the wave measured by two stationary observers at different distance from the star.

Gravitational redshift cannot explain the redshift of galaxies because if those galaxies actually didn't have redshift for observer in free space outside of our galaxy, for observer on earth they would have had a BLUE shift (the light sinks into a lower gravitational potential and increases in locally measured energy). Also the galaxies redshift has wide varying magnitudes that a single gravitational redshift can't encompass.

7. Jul 18, 2007

### duordi

So you are saying that the variation in red shift for all galaxies would be about the same if it were caused by a static gravitational field, I agree.

My question has to do with change in a gravitational field.

If my galaxy measuring stick is shrinking due to variation in time passage that would make the universe appear to stretch out with the farthest objects moving away faster then the close objects.

If my entire galaxie was shrinking with me, I might not be able to tell.

Or did you intend to say that the galaxies at a common distance have to large of a velocity differential to be caused by a variation in our condition?

8. Jul 18, 2007

### Voltage

I'm not quite clear on the above, but if I can offer a contribution...

smallphi: I understood that radial length contraction was a feature of General Relativity.

duordi: it might be useful if you considered your scenario where it was you receding from your mass at .99c.

9. Jul 18, 2007

### pervect

Staff Emeritus
If you want to tackle this problem from the perspective of GR, you need to have a metric. You can easily find a metric for a single massive body in a coordinate system where the massive body is at rest. (In fact there are several, the Schwarzschild metric is the most well known, but one could use the Painleve metric or isotropic coordinates).

Unfortunately, it is not easy to find a metric for a single massive body from some hypothetical POV where the massive body is moving. I'm not aware of any, and it's not possible to do any simple procedure to obtain them.

This is not a problem in getting results, as the paths of the light rays, and their energy-momentum 4 vector (which gives the redshift) can be calculated using a coordinate system in which the massive body is at rest.

To quote from http://arxiv.org/PS_cache/gr-qc/pdf/9508/9508043v1.pdf

So,the fundamental problem you are having is that you are trying to force-fit GR into a mode where it has an observer. The approach that will actually get you answers will require you to separate "the conceptual model of what is going on", which it turns out can be described concisely by the metric, from the notion of "an observer".

In a nutshell: the approach that will get you answers separates coordinates from "an observer". The widely known choices of coordinates are those that have the massive body at rest in their center.

The process of turning the coordinates into what you measure can be greatly aided by the conept of "frame fields", but this takes some knowledge of tensor calculus to appreciate.

So the good news is that we can compute answers to your questions, the bad news is that it doesn't use the "framework" that you want to impose, one of "an observer".

10. Jul 18, 2007

### daniel_i_l

You can look at the redshift this way:
Think of someone next to a star sending off light pulses with constant intervals into space. If an observer stationary relavtive to the star intercepts it he'll measure a larger time interval between the pulses.
If we interchange light pulses and peaks of a light wave the the time interval between the light pulses is like the frequency of the light.
So the observer measures a larger frequency - redshift.
Now going back to the pulses - if the observer is moving relative to the star then what happens? We can find out by first using a stationary observer in the same position as the moving one to measure the interval between the pulses, and then use SR to calculate the time interval measured by the observer moving relative to the stationary (relative to the star) observer. So the moving observer measures a slightly smaller interval than the stationary one.
If you want to include the moving observers mass then you use his rest mass. Really it's better to only use rest mass - "relativistic mass" just confuses things.
But I'm pretty sure (i didn't calculate it) that you'll get the same results whether you correct for time dialation or pretend that the moving observer is stationary but with a bigger mass.

11. Jul 18, 2007

### meopemuk

Yes.

Yes.

No. The gravitational red shift is fully determined by the rest mass of the source.

12. Jul 19, 2007

### duordi

Does everyone agree with meopemuk?

If so I have my answer.

The relativistic mass component does not add to gravitationaly imposed red shift.

13. Jul 19, 2007

### pmb_phy

Hmmmm! You mean to tell me that there was a relativistic mass thread going around an I didn't know about it/participate in it? That's way cool! Its finally nice to be able to ignore stuff like this.

Best regards

Pete

14. Jul 19, 2007

### duordi

Does anyone have a referance to the following statement being tested or proven by observation? I could not find one.

"The relativistic mass component does not add to gravitationaly imposed red shift."

Duane

15. Jul 19, 2007

### Voltage

Gravitational Lensing, duordi.

A photon has relativistic mass but no rest mass. It travels in a straight line until it skims a black hole or some other massive object. But its path is then bent by this object. Whilst the object is very massive compared to the photon, we know from Einstein that the object is also attracted towards the photon. And yet the photon only has relativistic mass. It isn't rest mass (aka invariant mass aka proper mass) that causes gravity. It's matter/energy that causes gravity. And matter is locked-up energy.

It's energy that causes gravity. And relativistic mass is considered to be a way of talking about energy.

IMHO.

16. Jul 21, 2007

### Schrodinger's Dog

No only reams of information that imply the opposite I'm afraid. If Gravity does not red shift or blue shift light then Einstein's theory of general relativity is apparently wrong.

From the wiki link given earlier.

If wiki is not your bag

http://physics.ucr.edu/~wudka/Physics7/Notes_www/node99.html

Last edited: Jul 21, 2007
17. Jul 21, 2007

### duordi

Voltage, your comment just counterdicted the statement by meopemuk above.

"The gravitational red shift is fully determined by the rest mass of the source."

Or at least I don't understand how they both can be true.

Duane

18. Jul 21, 2007

### Schrodinger's Dog

Correct me if I'm wrong but doesn't mass increase as you approach c, it's this that stops massive objects from achieving c, if that's the case would not the amount of red shift be effected by an objects speed, not just because of special relativity, but also because the actual objects mass would be increasing too? Or does an objects mass increase make no difference to red shift? If it does then we must allow that both special and general relativity indicate a red shift according to a masses speed, in which case the supposition that an objects rest mass determines it's red shift is a little contradictory, no? Or do they mean you can use the rest mass to determine red shift at any velocity in GR?

Last edited: Jul 21, 2007
19. Jul 21, 2007

### pervect

Staff Emeritus
This thread is starting to "drift" with rather speculative answers being given, I suppose I've been neglecting it.

For purely radial paths for the light, smallphi in post #9 and my post #6 have outlined the correct approach to calculating red shift. This does not involve using the concept of "relativisitc mass" at all, one performs all calculations in the frame of the massive body where one knows (can easily look up) the metric.

In a simple case where the geometry is entirely radial, one can multiply the gravitational redshift calculated for a stationary body by the SR doppler shift due to the relative velocity to get the total redshift.

This takes a bit more work to make precise - the velocity in this calculation is computed by measuring the velocity between the moving body and a "stationary" body at the same location. The "distance" in the calculation (needed to compute the gravitational redshift) is not a physical distance, but a Schwarzschild r coordinate. This takes advantage of the fact that in a static gravitational field there is a well-defined notion of "stationary", note that this is not true in more general case.

In more general cases with non-radial geodesics the multiplicative rule may not work (I'm not sure) - I believe the correct approach would involve considering the complete set of paths that light takes (technically called a null geodesic congruence). MTW takes something similar to this approach when they calculate the redshift due to the FRW metric on pg 777, the calculation is somewhat detailed and involves considering the proper time of the transmitter, the complete propagation path (null geodesic) of the light, and the proper time of the reciever at which the light signals are recieved.

You won't find much to answer this question as originally stated in the literature because nobody in GR, as far as I know, uses "relativistic mass". It's a bit of a dinosaur. Instead, they use one of several other concepts of mass, such as ADM, Bondi, or Komar mass, none of which is the same as the SR concept of mass, when they use mass at all.

20. Jul 21, 2007

### Voltage

Not quite, Schrodinger's Dog. "Mass" is nowadays defined to be invariant mass, commonly known as "rest mass", also known as "proper" mass, which is always the same irrespective of your reference frame. Relativistic mass increases as you approach c, this is a measure of energy. It's arguably the same thing as inertial mass and active gravitational mass, but these are defined differently. But it is confusing, check with pmb or pervect in case I've dropped a stitch.

duordi: I didn't mean to contradict meopemuk. There's plenty of room for confusion here, especially with the talk of a receding mass and redshift. If I can clarify my position, note my paragraph above. Gravitational mass is a measure of how much gravity an object "causes". Subject to differences in definition, it's the same as inertial mass, which is a measure of how much force is required to accelerate an object, and is the same as relativistic mass which is the same as energy. It isn't the same as rest mass. The rest mass doesn't "cause" the gravity, the energy does.

Consider a different scenario. You are in space close to a very very long stationary rod. You measure some amount of gravitational attraction caused by this rod. If however the rod was moving past you at some relativistic speed, you would measure more gravitational attraction caused by the rod.

Edit: pervect, we overlappped. I hope the above sounds reasonable.

Last edited: Jul 21, 2007
21. Jul 21, 2007

### meopemuk

Red shift is a frequency shift of spectral lines of atoms, which are placed in the gravitational field of a large mass M (Sun, Earth, Saturn, ...). If the frequency of radiation emitted by a free atom is $f$, then the frequency of radiation emitted by the same atom in the field is

$$f' = f (1 + \phi/c^2) [/itex] (1) where $\phi$ is the gravitational potential at the atom's location. For example, the gravitational potential created by the Sun is [tex] \phi = -\frac{GM}{r}$$ (2)

where G is the gravitational constant and r is the distance atom-Sun center. If your proposal is correct (the gravitational potential depends on velocity), then we would need to rewrite eq. (2) as

$$\phi = -\frac{GM}{r \sqrt{1 - v^2/c^2} }$$ (3)

where v is Sun's velocity with respect to the observer (us). In the $c^{-2}$ approximation

$$\phi = -\frac{GM}{r} (1 + \frac{v^2}{2c^2})$$ (4)

Then formula for the redshift (1) will take the form

$$f' = f (1 - \frac{GM}{rc^2} - \frac{GMv^2}{rc^4}) [/itex] So, your proposal results in corrections of the order $c^{-4}$ to the usual formula [tex] f' = f (1 - \frac{GM}{rc^2} )[/itex] which has been confirmed in experiments. These corrections are too small to be observed in modern measurements of red shifts. However, it *is* possible to reject your proposal on the basis of existing observations. If your formula (4) is correct, then the interaction energy in the system Sun-Mercury will be [tex] V = -\frac{GMm}{r} - \frac{GMmv^2}{2c^2r}$$

where $m$ is Mercury's rest mass. The $c^{-2}$ correction to the Newtonian interaction will lead to some observable effects on the Mercury's orbit, and one can show that these effects will be different from the perihelion precession that is actually observed.

Eugene.

Last edited: Jul 21, 2007
22. Jul 21, 2007

### Schrodinger's Dog

No that's ok, Pervect has cleared it up rather neatly Voltage thanks. it makes more sense when put in context rather than just the statement "The gravitational red shift is fully determined by the rest mass of the source."

Last edited: Jul 21, 2007
23. Jul 21, 2007

### pervect

Staff Emeritus
I think that there may be an alternate approach than the one I described earlier for calculating the redshift which is easier than considering the null geodesic congruences. But it still involves considering the massive object to be stationary. As I've mentioned before, the reason for doing this is that we can then use the well-known Schwarzschild metric to describe the metric of space-time. This is part of the abstraction process I mentioned, whereby we regard the metric as a complete specification of the physics, which we then use to calculate what individual observers "see" by means of frame-fields.

The basic idea is that we can model a photon as a particle with some energy-momentum 4-vector.

Using this approach, though, requires one to have enough knowledge of tensors to deal with covariant and contravariant components.

I'll sketch it out below - as far as relativistic mass goes, I'll say that I don't use it in the calculations, and that I've never seen a GR textbook use it either.

Using this approach, we can derive the redshift for a moving observer as follows:

1) We sepcify the energy E of the photon in some "frame-field" of the source. A frame field is just a local (not global) frame with a locally Minkowskian metric, i.e. a Minkowskian metric just like that of special relativity.

We can then write the energy in this frame as E = $\sqrt{|E^0 E_0|}$, where $E^0$ and $E_0$ are the covariant and contravariant components of the energy in Schwarzschild coordinates.

MTW refers to E as $E_{local}$.

2) In the Schwarzschild geometry, we know that $E_0$ is a conserved quantity of motion, also called the "energy at infinity". We can further simplify the problem by assuming that all motion is radial, so that the angular momentum of the photon is zero. In this case, specifying the energy of the photon and that it is outgoing is enough to specify the complete energy-momentum 4-vector.

3) Since the Schwarzschild metric is diagonal
$$E_{xmit} = \sqrt{|E^0 E_0|} = \sqrt{|g^{00} E_0 E_0|} = \sqrt{|g^{00}|} E_0 = \frac{E_0}{\sqrt{1-\frac{2 r_s}{r_{xmit}}}}$$

where $r_{xmit}$ is the Schwarzschild coordinate of the transmitter, and $r_s$ is the Schwarzschild radius of the large mass.

4) We can compute the energy-momentum 4-vector for a stationary (in Schwarzschild coordinates) receiver at some height $r_{rcv}$ by

$$E_{rcv} = \sqrt{|g^{00}|} E_0$$

because $E_0$ is a conserved quantity for the Schwarzschild geometry. This gives the well-known result

$$E_{rcv} = E_{xmit} \sqrt{\frac{1 - \frac{2 r_s}{r_{xmit}}}{1- \frac{2 r_s}{r_{rcv}}}}$$

Because frequency and energy are proportional, to get the energy for the moving reciever, we can apply the relativistic doppler shift, because our reciever is in a locally Minkowskian metric.

Thus we take the above forumla, and multiply it by the relativistic doppler shift factor

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/reldop2.html

of $$\sqrt{\frac{1+ \beta}{1 - \beta}}$$

Here $\beta$ is the velocity measured between a "stationary" observer at constant height $r_{rcv}$ and our hypothetical moving reciever in the frame-field of the receiver (or equivalently the frame-field of the moving observer).

We can perform this Lorentz boost because near the reciever, we can ignore the curvature of space-time, just as we can ignore the curvature of the Earth when dealing with nearby objects. Thus we can use SR techniques such as a Lorentz boost to convert from the frame-field of the reciever moving relative to the large mass to the frame field stationary relative to the large mass, and vica-versa, and we can also use the standard SR doppler shift formula. To make this work, it is important that we measure the velocity locally as well (i.e. either the moving observer or the stationary observer measures the velocity of the other while they are located at the same point, or very close to one another).

Last edited: Jul 21, 2007
24. Jul 21, 2007

### pmb_phy

relativistic mass and energy don't always have the same value. If the two were to be calculated given the stress-energy-momentum tensor then the energy-density would have a different value than the mass-density. If a person already believe that relativistic mass is always proportional to energy then it would seem predictable that the same person would always set them to be equal. This is a little known fact in relativity but a solidly known fact in the relativity literature. Take a look at this web page I wrote up to demonstrate this fact and why it is so.

http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

Schutz also discusses this in his new text "Gravity from the Ground Up."

Pete

25. Jul 22, 2007

### Voltage

I think I could enjoy talking to you about the finer points here Pete, and getting to the bottom of ρ0, but perhaps it's getting a little off topic. I'll make a note and check out that Schultz and take care with my words. Thanks for the input.

SD: Sorry, it's sometimes hard to know how to pitch a post, especially when pervect whacks one in whilst you're doing it. I'll bow out in case I muddy duordi's waters.