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Simple math question regarding Lorentz Invariance

  1. Dec 8, 2008 #1
    Let us restrict ourselves to SR for the moment at least. So we have a flat spacetime.
    Now consider a proper force of the form:
    [tex]\frac{dp^\mu}{d\tau} = a v^\mu[/tex]
    where [itex]a[/itex] is a scalar.

    It seems to be coordinate system independent due to the definition being in tensor notation. But it seems to not have any of the poincare symmetries. How do I show this explicitly?

    I'm confused since if I did a Lorentz transformation, the form is still the same, but it doesn't seem to have any conserved quantities. I know I'm confusing something very simple here. Can you help?
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  3. Dec 8, 2008 #2

    Jonathan Scott

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    The physics is of course unusual, in that your "proper force" appears to describe motion with constant four-velocity but with rest mass increasing at a steady rate.

    I don't get what you're worried about with the mathematical side of this question, but I may have missed something, as although I think I remember what "Poincare symmetries" are, I don't know how you would expect to show their presence or absence. As far as I can see, both sides are four-vectors with invariant scalar magnitudes. What do you mean about "conserved" quantities?
  4. Dec 8, 2008 #3
    I'm sorry the question isn't very clear. That is entirely my fault, because I am confused enough that I can't articulate very well what is confusing me.

    Let me try a specific example though.
    Choose an inertial coordinate system in which a particle feeling this force is at rest. The particle will remain at rest. Now choose a different inertial coordinate system in which the particle is moving. The particle's coordinate velocity will change.

    These are two different physical outcomes. One claims the rest frame is inertial, the other claims it is not. What is going on?

    It seems like this force somehow isn't coordinate system independent, even though it appears to be manifestly so. I am very confused. I'm hoping I'm making a simple (although probably fundemental) mistake that I can learn from here. I just have no idea where it is...
  5. Dec 8, 2008 #4

    Jonathan Scott

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    As I already mentioned, that particular force law does NOT change the coordinate velocity. It only changes the rest mass!
  6. Dec 8, 2008 #5


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    I'm not sure if it's helpful, but it's my understanding that any laws of physics--including, say, Newtonian gravity--will obey the same equations in any coordinate system (diffeomorphism invariance) if the laws are expressed in tensor form. See my post #8 on this thread. So, for a given law to be Lorentz-invariant I think you have to show it obeys the same equation in different inertial frames in SR when the equation is expressed in non tensor form.
  7. Dec 8, 2008 #6
    Oh! Okay. I see it now. I feel stupid.
    Thanks for explaining.

    If I apply a lorentz transformation, or a rotation, or any translation, the force law has the same form. But I don't see how any of the conserved quanties like energy or momentum or angular momentum, etc. that should arise when there are such symmetries can possible arise here.

    I'm still missing something important.
    Last edited: Dec 8, 2008
  8. Dec 8, 2008 #7
    Wait, I'm confused again. Unless a force law refers to some 'background field', if it is written in tensor notation, how could it possibly be non-lorentz invariant?

    Can you give an example for me to play with and learn?

    In this case, the "force" law can be rewritten as:
    [tex]\frac{d m}{d\tau} = \mbox{constant}[/tex]
    There are no indices even left. This is exactly what it will look like in any coordinate system, inertial or not.

    EDIT: Wait, I just reread your post. Coordinate independence yields diffeomorphism invariance? I thought diffeomorphism invariance was a stronger statement that coordinate independence. I'll read the links you gave. Maybe it will make more sense then.
  9. Dec 8, 2008 #8

    Jonathan Scott

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    In this case, the invariant quantity is the rate of change of the rest mass with respect to proper time (which for more conventional forces is normally zero).
  10. Dec 8, 2008 #9
    But where did all the other conserved quantities go?

    I think I may see the problem now. The action must be invariant to translations,rotations,etc. in order for those conserved quantities to arise. And due to explicit velocity dependence, it is not easy (impossible?) to write this in terms of an action. Is that right?

    So my (tentative) lesson here is: the symmetries of the force don't necessarily mean the action will have the same symmetries.

    If I'm going down the wrong 'road' here, please do let me know.
  11. Dec 8, 2008 #10


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    Do you mean a metric? Maybe "tensor notation" was too vague, I meant that as long as the force law is written in a tensor form that makes use of a metric tensor, then my understanding is that it has the same tensor equation in every coordinate system.
    My point is that an force law can be "non-Lorentz invariant" even though when it's written as a tensor equation using a metric tensor, it obeys the same equation in the different inertial coordinate systems related by the Lorentz transformation. To test if something is Lorentz-invariant or not you need to write the equation without a metric tensor.
    I don't have much understanding of tensor mathematics, I was just going by the stuff I read that I quoted in the other post.
    What do you mean "what it will look like in any coordinate system"? You can't just declare that this law applies in every coordinate system, writing the same equation in different coordinate systems can describe physically different force laws. In order to make sure you are talking about the same physical force law expressed in multiple coordinate systems, you have to take the equation in one coordinate system, then apply a coordinate transformation (like the Lorentz transformation) to this equation to find what the same physical force law would look like in a different coordinate system.

    Suppose we are talking about an inertial frame that uses x, t coordinates. How do you want the force law to be expressed in these coordinates? Is your [tex]\tau[/tex] above supposed to represent the time-coordinate of the frame we're using, or is it proper time? If the latter you have to re-express it using the frame's coordinates. Likewise, what is m supposed to represent?
  12. Dec 8, 2008 #11
    No, that is not what I meant by a background field. A background field is something associated with the vaccuum that can break Lorentz symmetry as it in essence picks a preferred frame. Besides a background field, I don't know of anything that can make a law formulated in tensor notation non-lorentz invarient.

    If for ease of showing my point, let's define an inertial frame by a coordinate system in which the the metric is diagonal -1,1,1,1. Therefore they are related by lorentz transformations. Since it is already defined how to transform the components of a particular coordinate system representation of a tensor (vector, scalar, etc.) to another coordinate system, all the vectors and tensors and scalars MUST transform from one inertial frame to another inertial frame according to the usual Lorentz transformations prescriptions.

    Therefore ANY physical law written in tensor form (even if you expand and write it in the components explicitly as you are demanding below) will transform correctly via the Lorentz transformations between inertial frames. The only thing that can break Lorentz invariance is if there is some background field (vector, tensor, etc.) associated with the vaccuum. That is because the vaccuum now has a 'preferred frame'.

    This doesn't make sense at all. Expanding out the equations in terms of the components doesn't change the form of anything. The only way what you wrote there could make sense is if you think the metric tensor can have a different coordinate representation in different inertial frames ... but it can't.

    The equation [itex]\frac{dm}{d\tau} = \mbox{constant}[/itex] is written in tensor notation. Notice there are no indices. That means everything there is a scalar. It looks the same in every coordinate system trivially, because everything is a scalar... there's nothing to even transform.

    I hope I didn't just completely misunderstand the point of your post.
    If so, I'm sorry for being so dense, but please do try again.
    Last edited: Dec 8, 2008
  13. Dec 8, 2008 #12


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    You're missing my point, which is that I think the term "Lorentz invariant" is defined in terms of how the equation transforms when you write it in non tensor notation. So your comment "I don't know of anything that can make a law formulated in tensor notation non-lorentz invariant" doesn't make sense--if the law is non-Lorentz invariant when you write it in non-tensor form, then that means it's a non-Lorentz invariant law, by definition, it doesn't matter that when you express it in tensor form it looks the same in every coordinate system. Now, I admit I'm not positive that this is how Lorentz invariance is defined since I'm not an expert in these things, but I take it as the implication of the stuff I quoted on the other thread about how any law, even something like Newtonian gravity, obeys the same tensor equations in every coordinate system--despite this, clearly no one defines Newtonian gravity as a Lorentz-invariant theory, because when you express it in non-tensor form and apply a Lorentz transformation to the equation, the equation doesn't remain the same.
    What do you mean by "transform correctly"? Of course if you know the equation in one frame, you can find the corresponding equation in another frame using the Lorentz transformation ('corresponding' meaning that if you use the new equation in the new frame, all your physical predictions about coordinate-independent facts like what various clocks read when they meet each other will be the same as when you used the original equation in the original frame). The question is whether the new equation is the same as the old equation, not in the sense of giving the same physical predictions but in the sense of having the same form when written down. For example, if we know the equation for Newtonian gravity works in one frame, so in the first frame an object's acceleration as a function of its position (with the position of the gravitating source being (x0,y0,z0)) is a(x,y,z) = GM/sqrt[(x - x0)^2 + (y - y0)^2 + (z - z0)^2], then if we apply the Galilei transformation to this equation, we get exactly the same equation but with x,y,z,and t replaced by x',y',z', and t', i.e. a'(x',y',z') = GM/sqrt[(x' - x0')^2 + (y' - y0')^2 + (z' - z0')^2]. On the other hand, if you apply the Lorentz transformation to the equation in the first frame, the corresponding equation you'd get in the new frame would not look just like the equation in the first frame but with x,y,z,t replaced by x',y',z',t'. So, Newtonian gravitation is Galilei-symmetric but not Lorentz-symmetric.
    I don't know quite what you mean by "change the form of anything". If you write down various equations in non-tensor form, there are some (like Maxwell's laws) where if you apply a Lorentz transformation you'll get back exactly the same equation but with x,y,z,t replaced by x',y',z',t', while there are others (like Newtonian gravity) where if you apply a Lorentz transformation you'll get a different equation in the new frame.
    I don't understand how what I said would imply I think the metric tensor would have a different representation in different inertial frames.
    But what I'm saying is that I think you have to put it in non tensor notation to test if it's Lorentz-invariant, the fact that a given law is unchanged under the the Lorentz transformation when expressed in tensor form does not mean the law is a Lorentz-invariant one, in my understanding (since the links I posted on the other thread indicated that any law would have this property, even Newtonian gravity which clearly is not defined as Lorentz-invariant).

    And again, what physical quantities are m and [tex]\tau[/tex] meant to represent in your equation?
    Last edited: Dec 8, 2008
  14. Dec 8, 2008 #13
    Since everything is a scalar in that equation, it is already in non tensor notation. That is the component representation in every coordinate system, including inertial coordinate systems.

    For example, consider the dimensionless coupling constant [itex]\alpha[/itex] from electrodynamics. If it appears in a tensor equation, how do you propose I rewrite it in a non-tensor equation. You don't need to change it, because it is a scalar. A scalar is already in its "coordinate representation".

    Much of what you you wrote in the last post seems to betray a misunderstanding of what a tensor equation is. If I show you a specific example of a four-vector, then by definition it transforms just like any other four-vector. Why do you think if I wrote out all the components of a tensor equation that transforming the components would be any different than transforming the four-vector and checking the component afterward. It is as if you think the tensor notation is not always equivalent to the component notation. Is that what you think?

    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. In doing so you will either have added a background field which breaks Lorentz symmetry, or made the theory Lorentz symmetric.

    Try to write Newton's gravity in tensor form and I think you'll immediately see what I mean.

    Newton's gravity can be written in terms of a potential giving (in appropriate units):
    [tex]\nabla^2 \Phi = 4 \pi G \rho[/tex]
    where rho is the mass density.
    Extending to tensor notation, (or at least the simplest extension) is:
    [tex]g^{\mu\nu}\partial_\mu \partial_\nu \Phi = 4 \pi G {T^{\alpha}}_{\alpha}[/tex]
    where T is the stress energy tensor.
    Many scientists thought this was the relativistic form of gravity till GR was proven experimentally, because this form IS lorentz invariant.

    That is not a fair example because what you started with is not a valid tensor equation. Your statement was that "My point is that an force law can be "non-Lorentz invariant" even though when it's written as a tensor equation using a metric tensor, it obeys the same equation in the different inertial coordinate systems related by the Lorentz transformation."

    Because if you expand out a tensor equation into its coordinate representation components, you still have the same equation. Are you going to transform a position four-vector and then claim the new components aren't the new positions in the new frame? I assume not, and similarly for any vector or tensor. However you keep bringing up that we need to include the metric tensor in the equations, and admittedly this doesn't correspond directly to a physical quantity, and the components of this tensor are just 'hidden constant' in the non-tensor notation of force laws. For example: [itex]F = q(E + v \times B)[/itex] we don't see the metric components explicitly.

    But we know the metric looks the same in all inertial coordinate systems. (You can transform it by hand to convince yourself of this.) So those constants coming from the metric will be the same in the new frame as well. Everything transfers over fine.

    The only way to break the Lorentz symmetry is to include some kind of background field that does it explicitly. For example a tensor that is diagonal 1,2,3,4 in one inertial frame, and is not associated with anything physical ... just the 'vaccuum'. If this was used in any force law, that tensor components would not look the same in other inertial frames and therefore break Lorentz symmetry.

    If your force law only involves the four momentums, four velocities, stress energy tensor, etc. of the fields and objects interacting ... along with the metric ... the theory has to be lorentz symmetric. Expanding then transforming the components individually is no different than transforming the vectors/tensors/etc and expanding later.

    EDIT: Oh, and m = invariant mass, and tau = proper time.
  15. Dec 8, 2008 #14


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    I'm not sure what the question is in this thread, but I'll give it a try anyway. :smile:

    Isn't this the source of the confusion? A particle that remains at rest in an inertial frame isn't feeling a force, by definition (of "force").

    You can define "proper acceleration" as the coordinate acceleration in a co-moving inertial frame, and then (I think) you can define the "force" in that frame as the (rest) mass times the acceleration. To be more precise, we define the force as the 4-vector that has components equal to [itex]md^2x^\mu/d(x^0)^2[/itex] in the co-moving inertial frame. I'm not sure how to use that to find the force at different points on the world line in one inertial frame. You'll probably have to use the definition at each point, and then transform all of the results to the frame you'd like to use.

    Another possible source of confusion is the old-fashioned definition of 4-vectors and tensors as "something that transforms as..." You can tell if "something" is a 4-vector according to that definition if and only if the definition of "something" specifies its components in all inertial frames. I know that this is obvious, but it's the sort of obvious statement that might be helpful (because the old-fashioned definition of tensors is so confusing that it's easy to overlook the obvious). One thing that's implied by this is that you can define a 4-vector by saying what its components are one inertial frame and saying that it's a 4-vector. The equation that you can use to test if "something" is a 4-vector then tells you the components of the 4-vector you're defining in all other inertial frames.

    Regarding the term "Lorentz invariant". I consider Lorentz invariance to be the same as "special covariance" which says that the laws of physics will "look the same" in all inertial frames when the components of the metric are replaced by the corresponding numbers.
  16. Dec 8, 2008 #15
    Discussions with Jonathan Scott helped me crystalize what was confusing me. The question is now basically here:
  17. Dec 8, 2008 #16


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    But if you have a coordinate system and you want the equations of motion in terms of that coordinate system, your equation doesn't tell you that, because it doesn't specify how the proper time is related to the space and time coordinates of your chosen coordinate system. What I'm saying is that an equation is Lorentz-invariant if, when you write it in terms of the coordinates of a single inertial frame in non-tensor form and then apply the Lorentz transformation, the resulting transformed equation looks exactly the same.
    Alpha just involves the ratio of various constants, it doesn't involve any variables that's can be expressed as a function of the x,y,z,t coordinates of a spacetime coordinate system. What I'm talking about is writing equations where any variables that are functions of the spacetime coordinates are written explicitly in terms of these coordinates.
    Can you specify what part of my post "seems to betray a misunderstanding of what a tensor equation is"? I don't see how anything I said would contradict the notion that all four-vectors transform the same way, for example.
    Do you agree that Newtonian gravity, when expressed in component notation, does not obey the same equation when the Lorentz transformation is applied? Do you agree that if we transform Newtonian gravity into a tensor equation involving a metric tensor, then it would obey the same tensor equation in all coordinate systems?
    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:
    Is the tensor equation above physically equivalent to the non-tensor form of Newtonian gravity? If so, then I doubt physicists would call it Lorentz-invariant--do you have a reference for the claim that they did?
    My point was that if you take a physical theory that in its non-tensor has different equations of motion in different inertial frames, then if you come up with a physically equivalent way of stating the theory using a tensor equation with a metric, then this tensor equation will apply in all inertial frames but that doesn't mean the theory is "Lorentz-invariant". I wasn't trying to come up with a demonstration that the tensor/metric form of ordinary Newtonian gravity would obey the same equation in different coordinate systems, because the math is beyond me, and because I already referenced an older thread containing links to sources which affirmed that any theory can be expressed in a coordinate-invariant tensor form. For example, here is one of the statements from my post on that thread:
    Am I incorrect in interpreting the quotes I posted to mean that "any physical theory" including Newtonian gravity can be expressed in terms of generally covariant equations which take an "identical form in all conceivable frames of reference"? If you don't disagree with this, then I shouldn't have to prove it explicitly in the case of Newtonian gravity.
    What do you mean by "same equation"? If the tensor equation involves a metric, and you want to rewrite it as a non-tensor equation with no metric that just gives equations of motion in terms of the spacetime coordinates (x(t), v(t), etc.), might not the equation look very different? For instance, if I am understanding the point about general covariance correctly, then you could express Maxwell's laws in an accelerated coordinate system like Rindler coordinates using exactly the same tensor equation that you'd use to express then in an inertial coordinate system. But clearly if we want to express Maxwell's laws in non-tensor form in terms of Rindler coordinates, the equations will be very different from the standard equations used to express Maxwell's laws in inertial coordinate systems.
    Would you agree that if we discovered that Newtonian gravity gave correct predictions in some inertial frame, and we wrote down Newtonian gravity in terms of a tensor equation with a metric, then the exact same metric equation would give correct predictions in any other inertial frame? Would you agree, though, that if Newtonian gravity gave correct predictions in some inertial frames, then the equations of motion in a different inertial frame would have to be different if they were expressed in non-tensor form? If you don't disagree with either of these then I don't really understand your point.
    But as I keep saying, I think you're using an incorrect definition of "Lorentz symmetric". If you can express a force law using a tensor equation involving a metric such that the equation works in every inertial frame, I'm saying that does not allow you to say the theory is Lorentz-symmetric. I'm saying it's part of the definition of Lorentz-symmetric that even if you express the theory purely in terms of an inertial frame's own coordinates with no metric involved, it still must obey the same equations in all the different inertial frames related by the Lorentz transformation. If it doesn't, then I think the theory is defined as non-Lorentz-symmetric, despite the fact that it does obey the same equations in every inertial frame if you use the form with a metric.
    Last edited: Dec 8, 2008
  18. Dec 9, 2008 #17
    But a scalar is a number that is the same in all coordinate systems. There is nothing to expand into variables. 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.

    Consider the scalar m (the rest mass). It can be written in terms of components if you wish:
    [tex]m = \sqrt{-p^\mu p_\mu}[/tex]
    Assuming the metric of an inertial coordinate system (ie. not explicitly denoting the metric pieces) we have:
    [tex]m = \sqrt{E^2/c^2-\vec{p} \cdot \vec{p}}[/tex]
    Is your worry that this won't look the same written in the components according to a different inertial frame? Because it HAS to look the same... that tensor equation holds in all frames, and expand in any inertial frame and you still get that SAME component representation.

    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. This is simply not possible for the reasons denoted above. If this is still not clear, I'm obviously not doing a good job explaining, but I'm not sure how to explain it any better since you seem to understand each piece but the whole result is just not 'clicking' for some reason.

    Sure, look at this:

    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 ... 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 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.

    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), a background field associated with nothing physical (or just the vaccuum) needs to be added in ... because then the 'background' (vaccuum) looks different to every inertial observer, and the theory is not Lorentz invariant.

    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. Unlike GR though, the spacetime doesn't 'couple' with energy and momentum. The spacetime is still just flat and a 'stage for physics' instead of an 'involved player' in the physics. 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. Look at the equation, what in there could possibly be non-Lorentz invariant? maybe this will help me understand better what is worrying you. The components of the vector or tensor MUST transform as you expect with the lorentz transformations between inertial coordinate systems, and the metric MUST be the same in all inertial coordinate systems, and scalars MUST be the same in all coordinate systems ... so what is in that equation that you are worried will not transform correctly?
  19. Dec 9, 2008 #18


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    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.
    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 [tex]\tau[/tex] 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.
    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).
    Again, I'm talking about how equations of motion transform. I had thought that your equation [tex]\frac{d m}{d\tau} = \mbox{constant}[/tex] 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 [tex]\frac{d m}{d\tau} = 0[/tex] by definition, and how does this tell you anything about the object's position as a function of time? Anyway, consider an equation like [tex]\frac{d\tau}{dt} = 0.6[/tex], 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 [tex]\frac{d\tau}{dt} = 0.6[/tex]? 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.
    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.
    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).
    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.
    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.
    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.
    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?
    Last edited: Dec 9, 2008
  20. Dec 9, 2008 #19


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    There's clearly some terminological confusion here between JesseM and JustinLevy, and I can't work my way through it all, but I can clarify one point.

    In the context of tensors, when people refer to a "scalar" they don't just mean any one-dimensional quantity, they mean an "invariant quantity" or a "rank (0,0) tensor" i.e. something that takes the same value in every coordinate system.

    That's something that used to confuse me when I first approached tensors.

    So invariant mass (rest mass), proper time, proper acceleration (magnitude) and the speed of light are all "scalars" in this sense, but coordinate time, coordinate distance and energy are not.
  21. Dec 9, 2008 #20


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    Thanks, I guess I was mistaken when I referred to [tex]\frac{d\tau}{dt}[/tex] as a "scalar" then. But in this case I'm confused about how you could express any equation of motion purely in terms of scalars--see my question to JustinLevy about the meaning of his equation [tex]\frac{d m}{d\tau} = \mbox{constant}[/tex]
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