# Why isn't there such a thing as an invariant velocity for all objects?

1. Apr 25, 2013

### Xilor

I don't know enough about general relativity to know why the following argument is wrong and I would love to hear why because I'm sure it will be a valuable lesson.

The path of light is influenced by gravitationally lensing.
Since the speed of light is invariant, the observed path of light can not depend on relative velocity differences between observer and light.
Since the speed of light is invariant, the observed path can not depend on lights relative speed to the mass it is lensed by.
The way that the path of light changes, and thus what the geometry of space is must be invariant under velocity of the observer.
The geometry of space is influenced by the massenergy of nearby objects. Their total massenergy must be invariant as well or there could not be invariant geometry.
Total massenergy includes rest mass, which is invariant under the velocity of the observer.
Total massenergy also includes kinetic energy, which must thus also be invariant under the velocity of the observer.
With kinetic energy and total mass energy known, you can calculate a velocity. This velocity must be invariant under the velocity of the observer.

2. Apr 25, 2013

### Bill_K

The speed of light is invariant, but the direction is not. The direction of a light ray will change depending on the motion of the observer. This effect is called aberration, and does not depend on relativity - it happens in nonrelativistic Newtonian physics also.

Energy also depends on the motion of the observer. It's just one part of a quantity called the stress-energy tensor, which is the real source of gravity in general relativity. Although the spacetime geometry is invariant in a sense, it appears different to observers that move through it differently.

Last edited: Apr 25, 2013
3. Apr 25, 2013

### Xilor

Well I know the direction changes through aberration, but that is an effect that seems to work completely differently.

Whether the geometry is invariant under the velocity of the observer as I would expect it to be should show from the answers to the following thought experiments. Which should have clear answers in relativity.

Lets take three objects in space, all separated in some axis X by a significant distance. The observer, the light source and an object in between causing light paths to change. Position and velocities in spatial dimensions Y and Z are 0 throughout the thought experiment. Initially. the observer and mass have no velocity difference between them along axis X either. The observer should now see light of the source lensed by the object in the middle to a circle around the object in the middle, a circle with some angular radius.

Lets say the mass in between stars starts moving along axis X, using some kinetic energy obtained from elsewhere. This increases the total energy of the object as measured, should change the path of light observed due to a different geometry and should widen the observed radius of the light from the source.

After that the observer accelerates to move equally along axis X. Will this revert the change and cause light to arrive like before? (ignoring the aberration effect)

That would sound very strange since it meant that the propagation of light was influenced by the movement of observer, long before it even arrived at the observer.

Taking it further, the light arriving right after acceleration of the observer was expected to have gone past the object in the middle a long time ago. Does this acceleration instantly cause the light to change?
What if after matching the velocity of the object in the middle, the observer stars accelerating in a direction again. No matter in which direction along axis X it accelerates, observed velocity of the object in the middle will increase. So should the observed light radius grow no matter in which direction the observer accelerates? Even if that acceleration takes place right the instant before light hits the observer?

If the answer to all of those questions is yes, shouldn't that be an observable effect?

If the answer to any of those questions is no, shouldn't that mean that the kinetic energy of objects is invariant under the movement of the observer? And thus that velocity is invariant?

4. Apr 25, 2013

### Bill_K

No, you seem to be saying that the light deflection past a moving mass M would be the same as the deflection past a somewhat greater mass M'. This is false. There's only one correct way to calculate the deflection: a) Lorentz transform to the rest frame of the mass, and use the standard deflection formula, then b) Lorentz transform back again. In step (b) the angle of the light path changes - due to aberration. You can't ignore aberration, because the change is entirely due to aberration, not kinetic energy.

Also, the "additional kinetic energy from elsewhere" argument does not work. If you import some rocket fuel or whatever, it has changed the total mass of the object, even while the mass is sitting still, before you fire the rocket.

Last edited: Apr 25, 2013
5. Apr 25, 2013

### Xilor

I see. Is that the case for all effects of general relativity? Where a moving mass M does not create equal effects to a slightly larger (but stationary) mass M'?

That does lead to some further questions.

Shouldn't the laws of physics uphold in all rest frames? Why give special worth to that rest frame of the mass for calculations? I can't see how the light would particularly care about that single rest frame nor how the observer would.

Even in that rest frame of the mass, kinetic energy from the temperature of the mass should still contribute right?
But shouldn't the total sum of effects come from the total sum of effects from each individual particle? Considering the effects of each individual particle from the rest frames of those individual particles, there would't be such a thing as kinetic energy from heat.

So why would it be ok to say that the observed kinetic energy from observer to mass does not play any role at all, while the kinetic energy from particles as seen from the rest frame of the mass does?

edit: Hmm and for the energy from elsewhere argument. How about filling space uniformly with gigantic space lasers that can accelerate the masses through photon pressure? Of course the lasers would keep the mass of space homogeneous by exchanging energy between them.

edit2: Weakly Interacting Massive Space Lasers

Last edited: Apr 25, 2013
6. Apr 25, 2013

### Staff: Mentor

There are probably some exceptions out there, but in general... yes, my interactions with something moving relative to me will be different from my interactions with something not moving relative to me. A smaller moving mass is not the same thing as larger stationary mass.

It doesn't, but it does care about the curvature of the spacetime that it's passing through, and that curvature is a lot easier to calculate in some frames so we do the calculations there.

Consider a more intuitive classical problem. I have a 1000kg elephant and a 10 gram bullet fired from an elephant gun. In the rest frame of the elephant, the bullet is moving at 1000 m/sec, the total momentum is 10 kgm/sec and the total kinetic energy 5x103 Joules. But in a frame in which the bullet is at rest, the elephant is moving at 1000 m/sec, the total momentum is 106 kgm/sec, and the total kinetic energy is 5x108. And it's exactly the same physics either way.
Energy and momentum are conserved in every frame, but they aren't necessarily the same in every frame, so if you're going to solve a problem you have to either work within a single frame or be very careful about how you transform from one frame to another.

7. Apr 25, 2013

### Xilor

I understand it is easier to calculate, but there should still be a way to calculate it in another rest frame right? That is definitely true in the bullet/elephant case.

Apparently a calculation in the observers reference frame would not have the kinetic energy of the mass as a variable. But a calculation in the reference frame of the mass does have kinetic energy of its parts as variables.
I can only think of one way to solve that, and that is by saying that movement energy is somehow different from temperature energy. Movement energy as it is relevant to gravitational effects would have to be considered invariant in that its effect is always 0. While temperature energy would be invariant while still being relevant.

And what then would you pick as a reference frame in say the lensing effects of a galaxy cluster? The movements of galaxies and stars within are really just temperature energy and part of the invariant mass. Do these movements have an effect or not?
If no, why then does the temperature of a single mass matter?
If yes, why did the individual motion of a mass not matter? If it is part of a galaxy it matters, but if it isn't then it doesn't?

8. Apr 25, 2013

### Staff: Mentor

I can always find a frame in which an object is at rest, and in that frame the object's kinetic energy will of course be zero. However, heat is sum of all the random movemenst of the individual atoms, and as they're all bouncing around in different directions I cannot find a frame in which they're all at rest; therefore I cannot find a frame in which the that energy vanishes.

9. Apr 25, 2013

### Xilor

Ok, but what does that mean for the galactic cluster example?

You have a star A with a certain velocity and invariant mass. From the perspective of the observer, the mass of star A matters for the observed angle of light arriving, the velocity does not. At no point in space would an observer be able to say that the velocity of that star mattered. Accelerate it in some direction and nothing changes. At no point in space would observers say that the acceleration of the mass changed the paths of light.

Now, the exact same star A is placed inside a galaxy cluster. Suddenly the velocity of star A matters because it is part of the invariant mass of the cluster. Accelerate star A in some direction and the angle of light will change for the observer. This last part would be true for all velocities and all possible masses of the galaxy cluster. An acceleration changes the invariant mass of the cluster. At all possible positions in space, observers would agree that the acceleration of that mass caused the angles of light to change.

How does that make any sense? What is the cluster doing to star A to make its velocity suddenly matter?

10. Apr 25, 2013

### Staff: Mentor

The cluster isn't doing anything to star A. It's doing something to the spacetime through which both star A and the light whose path we're calculating is moving.

Imagine a car on an infinite flat road. Because the road is perfectly flat, the car's zero-to-sixty time (speeds measured relative to the road by a clock at rest relative to the road) will be the same everywhere.

Suppose we consider the same car and the same road, except that now the road crosses a high mountain range so has some steep uphill and downhill sections. The zero-to-sixty time of the car will now vary depend on where we measure and which direction it's moving; but this is not because the mountains are doing something to the car, it's because they're doing something to the road.

11. Apr 25, 2013

### Xilor

Spacetime changes don't change the law of physics though. Lets give a complete example and then reduce it. I would love to hear where it is going wrong according to information told in this thread.

Consider two universes. Universes U and U'. Both contain exactly identical observers (telescopes on planets) and light sources. Universe B contains a star A and universe B' contains a star A'. Star A is moving at a velocity V relative to the observer, while star A' is moving at a different velocity V'. Except for that fact, universes U and U' are exactly equal. Stars A and A' have the same mass and are at the same distance from the observer at the moment of observation, the light sources are at the same distances and so on.

Stars A and A' affect the light deflection of their light sources to give images to the observers in their universes.

1. The universes remain empty other than the previously mentioned objects. Lets call them universes U1 and U1' for this step. The amount of deflection observed can be calculated. The velocities of the stars A1 and A1' are both 0 in their own reference frames and all other variables are equal as well, so the amount of deflection observed in universes U1 and U1' is the same.

2. Two universes U2 and U2' appear. They are exactly equal to U1 and U1' respectively, with the difference that universes U2 and U2' contain gigantic galaxy clusters G2 and G2'. Stars A2 and A2' are located inside these galaxy clusters. The different velocities of stars A2 and A2' mean that the average velocity of the galaxy clusters G2 and G2' is slightly different. The inertial mass of the galaxy clusters G and G' is also slightly different because the velocity difference between stars A2 and A2' will naturally be higher than the velocity differences between the average velocities of the larger systems G2 and G2'.
Since the inertial mass of G2 is different than the inertial mass of G2', the observer in U2 will see a different deflection of light than the observer in U2'

3. Universes U3 and U3' are duplicates of U2 and U2'. Except for the fact that the galaxies in both are reduced to single stars near A3 and A3' which are a light year away from them. The new two-star systems are called G3 and G3'.
The different velocities between A3 and A3' mean that the inertial masses of G3 and G3' are different. Since the inertial mass of G3 is different than the inertial mass of G3', the observer in U3 will see a different deflection of light than the observer in U3'

4. Universes U4 and U4' are duplicates of U3 and U3'. Except for the fact that the stars in both are replaced by a planets with telescopes, moved to the location of the observers. The new two-object systems are called G4 and G4'.
The different velocities between A4 and A4' mean that the inertial masses of G4 and G4' are different. Since the inertial mass of G4 is different than the inertial mass of G4', the observer in U4 will see a different deflection of light than the observer in U4'

So 1. must have been incorrect. Movement of mass A will influence the deflection of light and the geometry of space.

12. Apr 25, 2013

### Bill_K

This is already incorrect. As stated way back at the beginning, the light deflection is the same in the rest frame of the stars, A1 and A1'. However since they are moving at different velocities, the light rays received by the observer will have undergone different aberration, and the deflection measured by the observer will be different.

13. Apr 25, 2013

### Xilor

Hmm right, I forgot about that. I'm mostly just trying to look at it from an energy perspective though. But would you happen to have a link for further reading? Most of what I can find seems to just deal with differing observed directions of light at different velocities of the observers. Not anything that has to do with gravitational lensing and how the movement of the mass affects how it deflects light.