JustinLevy said:
But a scalar is a number that is the same in all coordinate systems. There is nothing to expand into variables.
It depends on the scalar. If you have a scalar that's defined as a ratio between two variables that are not dimensionless, and at least one of those variables can be expressed as a function of the spacetime coordinates of whatever coordinate system you're using, then here you obviously do have something you can expand into variables.
JustinLevy said:
If you know a relation between the scalar and some components (for instance the norm of a four vector can be written using the metric and the components if you wish), because the metric is the same in all inertial coordinate systems insisting on writing it out in components won't change anything ... the equation will still be the same function of the components of that four vector in any inertial frame.
When you say "writing it out in components", what does "it" refer to specifically? I'm not talking about stuff like how a four-vector might appear in a specific coordinate system, I'm talking about what happens when you write the
equations of motion in terms of the spacetime coordinates (x,y,z,t), with no variables like \tau that are functions of those coordinates. Do you agree that if you do this, the equations of motion for Newtonian gravity won't remain the same when you apply the Lorentz transformation? It would really help me if you would address this specific example, I've asked you about it repeatedly and you haven't answered.
JustinLevy said:
Consider the scalar m (the rest mass). It can be written in terms of components if you wish:
m = \sqrt{-p^\mu p_\mu}
Assuming the metric of an inertial coordinate system (ie. not explicitly denoting the metric pieces) we have:
m = \sqrt{E^2/c^2-\vec{p} \cdot \vec{p}}
Is your worry that this won't look the same written in the components according to a different inertial frame?
No, I'm talking specifically about how equations of motion transform when you write them out, writing m out in terms of components does not give an equation of motion. And in any case writing m in terms of E and p is not the sort of expansion I'm talking about, I'm only talking about the spacetime variables x,y,z,t (obviously you couldn't expand out a momentum component px in terms of the spacetime variables without including the constant rest mass m).
JustinLevy said:
I don't understand why you think a scalar, if we use a relation to rewrite it in terms of some components of a four vector or tensor, can possibly look different in different inertial frames.
Again, I'm talking about how equations of motion transform. I had thought that your equation \frac{d m}{d\tau} = \mbox{constant} was supposed to be some type of equation of motion, although thinking about it, I don't really understand it very well--if the rest mass m is constant then doesn't \frac{d m}{d\tau} = 0 by definition, and how does this tell you anything about the object's position as a function of time? Anyway, consider an equation like \frac{d\tau}{dt} = 0.6, which also involves only scalars. If there is an inertial frame where it's true as a general law that all particles of a certain type satisfy this equation, that must imply that these particles are all bound by the laws of nature to move at a constant speed of 0.8c in this frame. This still isn't really a full equation of motion because it doesn't tell you how an arbitrary set of initial conditions involving these particles would evolve (for example, although it says their speed must be constant it doesn't tell you if their direction must be constant as well), but suppose it could be derived as a consequence of some more general equations of motion for these particles. Would you agree that whatever the more general equations are, when you write them in terms of the x,y,z,t coordinates of different inertial frames, it cannot possibly be true in
all inertial frames that \frac{d\tau}{dt} = 0.6? After all, if the particles are bound by the laws of nature to move at 0.8c in one frame, in other frames their speed must be something different.
JustinLevy said:
Newton's law is not a tensor equation. To turn it into a tensor equation you'd have to add quite a bit ... in essence extending it to become a geometric theory.
JesseM said:
Doesn't that just mean using a metric tensor? If so, that's exactly why I clarified earlier that I was specifically talking about tensor equations that used a metric tensor
JustinLevy said:
It means a heck of a lot more than just throwing in a metric tensor somewhere.
You keep making comments as if you just add the metric tensor in places, and you have turned a theory into a geometric theory
I've said that I don't really understand tensor equations involving metrics, so it wouldn't surprise me if I still wasn't being specific enough with the phrase "tensor equations that used a metric tensor". But if you remember that all this was meant to be in reference to the stuff I posted on the other thread, I think my intent was clear enough, even if I got the details wrong: I'm basically just talking about "whatever it is you have to do to some equations of motion to convert them into a generally covariant form that will obey the same equations in all coordinate systems". If this requires something more than just writing the equations of motion as a tensor equation involving a metric, then fill in the blanks, I obviously have little knowledge of the details but the only thing relevant to my larger point is that there is
some way of doing this for any arbitrary laws of physics.
JustinLevy said:
and because you know the original theory was not Lorentz invariant, you seem to combine this misbelief to conclude this means the metric pieces somehow are what make the tensor equation the same in all coordinate systems but the component equation look different in inertial frames.
I have not used the phrase "component equation", and I am not sure if the way you are using it precisely corresponds to what I am talking about when I say we should write the equations of motion out in terms of the spacetime variables x,y,z,t. Let's go back to the example I keep bringing up, the hypothetical scenario where Newtonian gravity is precisely correct in some inertial frame. Apparently there is
some method of writing the laws of motion for this universe in a generally covariant form that will still be the same in other inertial frames--maybe it involves something more than just writing the equation with a metric, I don't know. But whatever this generally covariant form is, it presumably involves a metric (even if that is not a
sufficient condition to ensure that we have a generally covariant form of the laws of motion), no? Is there going to be some "component equation" for this generally covariant equation expressing Newtonian gravity? If so, are you saying the component equation would also be generally covariant, so it would look the same in every inertial frame? Clearly if we write the law of motion for Newtonian gravity out in terms of the spacetime variables the equation will
not look the same in every inertial frame (assuming we define the relationship between frames in terms of the Lorentz transformation rather than the Galilei transform).
JustinLevy said:
I think that is your key mistake here. You are not understanding what a tensor equation is. Or at least you are not understanding what the process of promoting a theory to a geometric theory involves.
No, I've said pretty explicitly that I don't understand the details of that, but all that matters to my argument is that there is
some way of promoting an arbitrary physical law into a generally covariant form that will obey the same equations in every coordinate system.
JustinLevy said:
To break Lorentz symmetry, instead of just adding in a metric (which is the same in all inertial frames, so can't possibly be the cause of breaking Lorentz invariance)
Nowhere have I said or implied that you need to add a metric to "break Lorentz symmetry". What I thought--and you seem to say this was a misunderstanding--was that if we have a theory like Newtonian gravity which is already non-Lorentz-symmetric, we can convert it into a generally covariant form that will obey the same equations in every frame by rewriting it in terms of a tensor equation that uses a metric. If this is not sufficient, please tell me what words I should use to describe what needs to be done to convert an arbitrary theory into a generally covariant form.
JustinLevy said:
Go back and look at that example extension of Newtonian gravity into a geometric theory. It is an example of a relativistic theory of gravity.
Obviously I don't understand enough about tensors and metrics to interpret the equation you posted, which is why I asked you some questions about it. Again, is it
physically equivalent to a theory that says Newtonian gravity works precisely in one special inertial frame? For example, is it true that there is a preferred frame where if a planet is accelerated by a non-gravitational force, all distant masses will react instantaneously to the planet's change in position according to that frame's definition of simultaneity? If so, then obviously since other inertial frames define simultaneity differently, they must see distant masses react to the acceleration either before or after the planet was accelerated. So presumably if different frames write out the equations of motion in terms of their own spacetime coordinates, the equations must be different in different frames, so the theory is non-Lorentz-invariant.
On the other hand if the equation you wrote down is
not physically equivalent to a theory that says Newtonian gravity works precisely in one inertial frame, but instead represents a theory where all inertial frames will see the equations of motion take the same form when written down as ordinary differential equations involving the spacetime coordinates, then I don't really see how it's relevant to the discussion.
JustinLevy said:
This is a SR theory of gravity. It clearly doesn't match experiment, but it reduces to Newtonian gravity in the right limit, and is lorentz invariant.
Since you say it "reduces to Newtonian gravity in the right limit" does that mean it isn't actually fully equivalent to Newtonian gravity in any inertial frame? If not, then again, I don't understand the relevance. My first point was about how a theory's being "non-Lorentz-invariant" means that when you write down the equations of motion in terms of each frame's own spacetime coordinates with no metric involved, the equations will look different in different frames. My second point was that if you have a theory that is non-Lorentz-invariant in this sense, then the fact that you can rewrite the exact same physical theory (not a new theory that reduces to it in certain limits) as a generally covariant equation that works the same in all inertial frames doesn't change the fact that the theory is Lorentz-invariant. If what you were presenting was a new theory of gravity that was
not physically equivalent to a theory where Newtonian gravity is precisely correct in one preferred inertial frame, and which
was Lorentz-invariant according to my first definition, then what does this have to do with discounting either of my points?
