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Lorentz transformation of y cpmponent for 4-momentum

  1. Dec 3, 2012 #1
    I have 2 coordinate systems which move along ##x,x'## axis. I have derived a Lorentz transformation for an ##x## component of momentum, which is one part of an 4-momentum vector ##p_\mu##. This is my derivation:

    p_x &= mv_x \gamma(v_x)\\
    p_x &= \frac{m (v_x'+u)}{\left(1+v_x' \frac{u}{c^2}\right) \sqrt{1 - \left(v_x' + u \right)^2 / c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2}} \\
    p_x &= \frac{m (v_x'+u) \left( 1+ v_x' \frac{u}{c^2} \right)}{\left(1+v_x' \frac{u}{c^2}\right) \sqrt{\left[c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2 - \left(v_x' + u \right)^2 \right] / c^2 }} \\
    p_x &= \frac{m (v_x'+u)}{\sqrt{\left[c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2 - \left(v_x' + u \right)^2 \right] / c^2 }} \\
    p_x &= \frac{m (v_x'+u)}{\sqrt{\left[c^2 \left( 1+ 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4} \right) - v_x'^2 - 2 v_x' u - u^2 \right] / c^2 }} \\
    p_x &= \frac{mv_x'+mu}{\sqrt{\left[c^2 + 2 v_x'u + v_x'^2 \frac{u^2}{c^2} - v_x'^2 - 2 v_x' u - u^2 \right] / c^2 }} \\
    p_x &= \frac{mv_x'+mu}{\sqrt{\left[c^2 + v_x'^2 \frac{u^2}{c^2} - v_x'^2 - u^2 \right] / c^2 }} \\
    p_x &= \frac{mv_x'+mu}{\sqrt{1 + v_x'^2 \frac{u^2}{c^4} - \frac{v_x'^2}{c^2} - \frac{u^2}{c^2} }} \\
    p_x &= \frac{mv_x'+mu}{\sqrt{\left(1 - \frac{u^2}{c^2}\right) \left(1-\frac{v_x'^2}{c^2} \right)}} \\
    p_x &= \gamma \left[mv_x' \gamma(v_x') + mu \gamma(v_x') \right] \\
    p_x &= \gamma \left[mv_x' \gamma(v_x') + \frac{mc^2 \gamma(v_x') u}{c^2} \right] \\
    p_x &= \gamma \left[p_x' + \frac{W'}{c^2} u\right]

    I tried to derive Lorentz transformation for momentum also in ##y## direction, but i can't seem to get relation ##p_y=p_y'## because in the end i can't get rid of ##2v_x'\frac{u}{c^2}## and ##\frac{v_y'^2}{c^2}##. Here is my attempt.

    p_y &= m v_y \gamma(v_y)\\
    p_y &= \frac{m v_y'}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{1 - v_y'^2/c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2}}\\
    p_y &= \frac{m v_y' \left( 1 + v_x' \frac{u}{c^2} \right)^2}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{\left[c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2 - v_y'^2\right]/c^2}}\\
    p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2 - v_y'^2\right]/c^2}}\\
    p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2\left( 1 + 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4}\right) - v_y'^2\right]/c^2}}\\
    p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2 + 2 v_x' u + v_x'^2 \frac{u^2}{c^2} - v_y'^2\right]/c^2}}\\
    p_y &= \frac{m v_y'}{\gamma \sqrt{1 + 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4} - \frac{v_y'^2}{c^2}}}\\

    This is where it ends for me and I would need someone to point me the way and show me, how i can i get ##p_y = p_y'##. I haven't seen any derivation like this (for ##y## component of momentum) on the internet.

    Thank you.
  2. jcsd
  3. Dec 3, 2012 #2


    Staff: Mentor

    This seems awfully complicated; plus, you are starting with an incorrect assumption about p_y.

    The simplest way to see how 4-momentum transforms is to realize that it is a 4-vector, just like the "position" (t, x, y, z). Its components are (E, p^x, p^y, p^z), and they transform the same way any other 4-vector does. That is, you have, for relative motion in the x direction (and in units where c = 1),

    [tex]E' = \gamma \left( E - v p^x \right)[/tex]

    [tex]p'^x = \gamma \left( p^x - v E \right)[/tex]

    [tex]p'^y = p^y[/tex]

    [tex]p'^z = p^z[/tex]

    Which corresponds to

    [tex]t' = \gamma \left( t - v x \right)[/tex]

    [tex]x' = \gamma \left( x - v t \right)[/tex]

    [tex]y' = y[/tex]

    [tex]z' = z[/tex]

    The transformation for the momentum 4-vector can be derived the same way the transformation for the position 4-vector is derived. The easiest way is to start with the invariance of rest mass: [itex]m^2 = E^2 - (p^x)^2 - (p^y)^2 - (p^z)^2 = E'^2 - (p'^x)^2 - (p'^y)^2 - (p'^z)^2[/itex], which corresponds to the invariance of the spacetime interval for the position 4-vector.
  4. Dec 3, 2012 #3


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    [itex]p_x = m v_x \gamma(v_x)[/itex] is wrong. It should be [itex]p_x = m v_x \gamma \left(\sqrt{v_x^2 + v_y^2 + v_z^2} \right)[/itex] (and similarly for the y and z components).
  5. Dec 3, 2012 #4
    How can this be wrong if i get a good result in my first derivation?

    I am sorry i can't just believe in a statement that 4-momentum transforms just like a spacetime. I would need a proof for this.
  6. Dec 3, 2012 #5


    Staff: Mentor

    Do you believe the last equation I wrote down in my last post? That's the standard energy-momentum relation in SR. Once you have that, deriving the transformation law for the components of 4-momentum is exactly the same as deriving the standard Lorentz transformation from the invariance of the spacetime interval.
  7. Dec 3, 2012 #6


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    [tex]p^\alpha = m\frac{dx^\alpha}{d\tau} = \frac{dt}{d\tau} \, m\frac{dx^\alpha}{dt} = \left( \gamma mc, \gamma mv^x, \gamma mv^y, \gamma mv^z\right)[/tex]
    [itex]dx^\alpha[/itex] transforms via the Lorentz transform. [itex]\tau[/itex] is invariant.

    All 4-vectors transform via the Lorentz transform; that's what makes them 4-vectors.
  8. Dec 4, 2012 #7
    I assume you mean:
    m^2 = E^2 - (p^x)^2 - (p^y)^2 - (p^z)^2 = E'^2 - (p'^x)^2 - (p'^y)^2 - (p'^z)^2

    Why should i believe that? How exactly does that prove ##p_y = p_y'## if relative speed ##u## among coordinate systems ##S## and ##S'## is in direction of ##x##, ##x'## axis? To translate momentum or energy in a different frame i refered to this site's section "Transforming Energy and Momentum to a New Frame".

    Following this sites sugesstions i have been able to derive 2 equations which are (and this is half of 4-momentum):
    p_x &= \gamma \left[p_x' + \frac{W'}{c^2} u\right]\\
    W &= \gamma \left[ W' + p' u \right]

    But when i tried deriving ##p_y = p_y'## or ##p_z = p_z'## i couldn't proove them. And this is weird. This method should work for ##p_y## and ##p_z## just as it worked fine for ##W## and ##p_x##.
    Last edited: Dec 4, 2012
  9. Dec 4, 2012 #8


    Staff: Mentor

    Because it's a way of expressing the invariance of rest mass, which is an important part of SR. Are you saying you don't believe SR? What, exactly, are the things you're willing to accept as true in order to start your derivation of the Lorentz transformation?

    Btw, if you look at the site you linked to, it has a formula that's equivalent to the one I gave: they write [itex]E^2 - c^2 p^2 = m_0^2 c^4[/itex], which is what I wrote if you adopt units in which c = 1 (and I wrote m instead of m_0), and recognize that this equation holds in every frame, so it holds for E' and p' as well as E and p (in fact they write this explicitly further down the page). In fact, the derivation they go on to do from this is exactly what I was talking about when I said you can derive the LT for 4-momentum from the invariance of rest mass.

    Yes, these look right, they're the same transformation equations that I wrote down, except that they write W instead of E and u instead of v.

    Why do you think that? The relative motion is in the x direction, so the x direction is different than the y and z directions. The y and z momentum does not change at all for relative motion in the x direction. The y and z *velocities* change, because they are affected by time dilation, but the *momentum* in the y and z directions doesn't change.
  10. Dec 4, 2012 #9


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    I'm not sure what kind of proof you are looking for. If for a slower-than-light particle, you define proper time [itex]\tau[/itex] to be the quantity [itex]\int \sqrt{1-\dfrac{v^2}{c^2}} dt[/itex], then you can prove that the quantity [itex]P^\mu = m \dfrac{dx^\mu}{d\tau}[/itex] is a 4-vector, since [itex]\tau[/itex] is a scalar. It's easy to prove that in the limit as [itex]v/c \rightarrow 0[/itex], the spatial part goes to the Newtonian momentum. Is it really momentum? Well, what's your definition of momentum?
  11. Dec 5, 2012 #10
    I presume derivation in my opening post is not going to work, so i would like to try it your way. I decided i will follow this site in combination with this and this video (this is for reference if anyone else will need it). Now that i decided to follow what all of you have been saying i stumbeled uppon a problem.

    You see our professor stated that invariant interval is ##\Delta s^2 = \Delta x^2 - (c \Delta t)^2## if we ommit dimensions ##y## and ##z##. Soo i presume for 4-D it wold be ##\Delta s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2 - (c \Delta t)^2##.

    Last equation for invariant interval isn't like on most sites as it has negative time component and positive space components while yours is vice versa. Why is that?

    How do i derive 4-momentum if i start with only 3 equations below.

    \Delta s^2 &= \Delta x^2 + \Delta y^2 + \Delta z^2 - (c \Delta t)^2\\
    p &= mv \gamma(v)\\
    E &= mc^2 \gamma(v) = E_k + mc^2

    I am sorry for such questions but our professor didn't use standard notaions and therefore i am having a hard time now.
  12. Dec 5, 2012 #11


    Staff: Mentor

    Yes, with the professor's sign convention (see below), this is correct.

    It's just a different sign convention. The professor's equation that you wrote above makes s^2 negative for a timelike interval and positive for a spacelike interval. The other sign convention makes s^2 positive for a timelike interval and negative for a spacelike interval. Some people prefer one, some the other. The physics is the same either way; you just have to keep track of which sign convention you're using and be consistent.

    We have

    E^2 = m^2 c^4 \gamma^2 \\
    p^2 = m^2 v^2 \gamma^2


    [tex]E^2 - p^2 c^2 = m^2 c^4 \gamma^2 \left( 1 - \frac{v^2}{c^2} \right) = m^2 c^4[/tex]

    which is the energy-momentum relation I wrote down earlier. If you want to expand out p^2 by components, you would have

    [tex]E^2 - p_x^2 c^2 - p_y^2 c^2 - p_z^2 c^2 = m^2 c^4[/tex]


    p_x = m v_x \gamma \\
    p_y = m v_y \gamma \\
    p_z = m v_z \gamma
  13. Dec 5, 2012 #12
    So what my professor does is correct? Please anwser me with YES/NO.
    What bothers me is that convention our professor used matches with hyperbola equations. And it seems to me yours doesn't. Let me explain why. First please take a look at the Minkowski diagram where there are 2 hyperbolas.

    http://shrani.si/t/2U/3Z/3ZCiUipK/capture.jpg [Broken]

    I have figured out that i could connect invariant interval to the semimajor axis of hyperbolas in the picture. I started out from basic hyperbola equation (i ll do this for hyperbola with ##a=b=2## in the picture - I can state this as asymptotes of "lightning cone or. X" are perpendicular to eachother).

    \frac{x^2}{a^2} - \frac{y^2}{b^2} &= 1\\
    \frac{x^2}{2^2} - \frac{y^2}{2^2} &= 1\\
    x^2 - y^2 &= 2^2\\

    And then i figured out that the axis we generally write ##y## is actually ##ct## axis in Minkowski diagram while axis we generally write ##x## stays the same in Minkowski diagram. So i get an equation, where quantity under the square root on the left hand side of an equation represents invariant interval ##\Delta s##.

    x^2 - (ct)^2 &= s^2\\
    \Delta s^2 &= \Delta x^2 - (c \Delta t)^2 \\

    This equation corresponds to the conventionbelow that my professor used.

    \Delta s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2 - (c \Delta t)^2

    I wonder what would change in my picture if i would use your convention below

    \Delta s^2 = - \Delta x^2 - \Delta y^2 - \Delta z^2 + (c \Delta t)^2

    I think your convention comes from different basic hyperbola equation ##\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1## and i would therefore get hyperbolas which open to the left/right instead of ones that open up/down? Please correct me if i am wrong. This is what i know about invariant interval and convention, but i don't know if i am correct.
    I understand this, but last equation is derived using your convention (scalar is positive and vectors all negative). How could i write it down for my convention? Is it like this?

    p^2 c^2 - E^2 = -m^2 c^4\\
    (pc)^2 - E^2 = -(mc^2)^2\\

    Well I can see that invariant is ##mc##, but it is negative! Is this ok? (here is allso negative) I ask this because in spacetime invariance ##\Delta s## was positive. If i divide above equation by ##c^2## i get

    p^2 - \frac{E^2}{c^2} = -(mc)^2\\

    At this point you would probably write left hand side of an equation using components, while you would state that right hand side is the dot product of an 4-momentum vector and therefore make a conclusion that 4-momentum vector is:

    p_x^2 + p_y^2 +p_z^2 - \frac{E^2}{c^2} = p^\mu \cdot p^\mu\Longrightarrow \boxed{p^\mu = (p_x, p_y, p_z, \frac{E}{c})}

    How do you know that ##p_x^2 + p_y^2 +p_z^2 - \frac{E^2}{c^2}## is a dot product of a 4-vector with itself?

    Should a 4-momentum vector be ##p^\mu = (p_x, p_y, p_z, -\frac{E}{c})## instead of ##p^\mu = (p_x, p_y, p_z, \frac{E}{c})## and why don't we usually write down a minus sign here?
    Last edited by a moderator: May 6, 2017
  14. Dec 5, 2012 #13


    Staff: Mentor

    [Edit: Added some clarifications.]


    There are two sets of hyperbolas. One set corresponds to a spacelike s^2--i.e., an interval where the spacelike components (x^2 + y^2 + z^2) are larger in magnitude than the timelike component (t^2). The hyperbolas you drew are from that set.

    The other set corresponds to a timelike s^2--i.e., an interval where the timelike component is larger in magnitude than the spacelike components. This set of hyperbolas would indeed, as you said, spread to the left/right instead of up/down.

    It's important to realize that the sign convention for s^2--i.e., whether you write the interval the way your professor did, with t^2 negative, or the way I did, with t^2 positive--is independent of the above distinction. You can write hyperbolas that spread left/right instead of up/down with the t^2 term negative; you just have to also write a negative s^2, which, as I said before, corresponds to a timelike interval with that sign convention. For example, considering just the t and x coordinates, we could write:

    [tex]x^2 - c^2 t^2 = - 1[/tex]

    which would correspond to a hyperbola spreading left/right. This isn't the normal way of writing hyperbolas in high school math class, but it works.

    You have just opened a different can of worms, but hopefully we can re-can them without too much trouble. :wink:

    Consider intervals first; as we've seen, there are three different kinds (though we haven't talked much about the third yet):

    (1) Timelike intervals, which have negative s^2 in your professor's sign convention and positive s^2 in mine;

    (2) Spacelike intervals, which have positive s^2 in your professor's sign convention and negative s^2 in mine;

    (3) Null intervals, which have s^2 = 0 (obviously the sign convention doesn't matter here).

    These three kinds of intervals describe three physically different kinds of things:

    Timelike intervals describe "lengths of time"--put another way, a curve with a timelike s^2 is a possible worldline for an ordinary observer with nonzero rest mass, and the length of the curve (i.e,. s) is the elapsed time experienced by the observer.

    Spacelike intervals describe ordinary "lengths in space"--put another way, a curve with a spacelike s^2 is a possible "curve in space at some instant of time" for some observer, and the length of the curve (s) is the distance measured by that observer.

    Null intervals describe light rays--a curve with null s^2 is a possible worldline for a light ray.

    Now consider the corresponding 3 kinds of 4-vectors with energy-momentum components:

    (1) Timelike 4-momentum, which has positive m^2 (I'll leave out the factors of c here, we're now working in units where c = 1) in my sign convention.

    (2) Spacelike 4-momentum, which has negative m^2 in my sign convention.

    (3) Null 4-momentum, which has zero m^2.

    A timelike 4-momentum describes the energy and momentum of a timelike object--i.e,. one with nonzero rest mass (the rest mass is just the length, m, of the 4-momentum vector) which moves on a worldline with a timelike s^2.

    A null 4-momentum describes the energy and momentum of a light ray, which moves on a worldline with null s^2.

    A spacelike 4-momentum would then describe a hypothetical "object" (the usual name for these objects is "tachyons") which moves on a spacelike worldline, i.e., one with spacelike s^2. You can find a *lot* of articles about tachyons by Googling, but for a quick overview I recommend the Usenet Physics FAQ's article:


    The fact that we normally view energy as a real number, with a positive square, is why we normally adopt my sign convention, with m^2 positive for timelike 4-momentum, when describing 4-momentum vectors, even when we are using your professor's convention for intervals (with s^2 negative for timelike intervals). And hopefully that gets the worms most of the way back into the can for now. :wink:

    I'm pressed for time right now so I'll defer responding to the two questions at the end of your post, since they raise some other issues we haven't touched on yet.
    Last edited: Dec 5, 2012
  15. Dec 5, 2012 #14


    Staff: Mentor

    Now for those two questions:

    Because the dot product of a vector with itself gives its squared length, and that's what you just wrote the formula for.

    You may be confused because you're used to seeing a dot product written with all plus signs, and there's that minus sign in front of E^2. The dot product you're used to seeing is for ordinary Euclidean space, where all squared lengths are positive. The more technical way of saying this is that the metric of ordinary Euclidean space is "positive definite": the squared length of any nonzero vector is a positive number. The "metric" in ordinary Euclidean space is just the Pythagorean theorem in three dimensions: [itex]s^2 = x^2 + y^2 + z^2[/itex]. And of course this is just the ordinary dot product of the vector (x, y, z) with itself.

    In spacetime, as we have seen, we can have nonzero vectors with positive, negative, or zero squared length. (Your professor's sign convention makes spacelike squared lengths positive, which is natural when you are thinking about the analogy with Euclidean space; that's why it's so common.) So the concept of "dot product" needs to be generalized to cover this case. The way we generalize it is simple: the dot product is computed using the metric, meaning the analogue of the Pythagorean formula for spacetime. So the interval we've been looking at, [itex]s^2 = x^2 + y^2 + z^2 - c^2 t^2[/itex], is just the dot product of the spacetime "position vector" with itself, using the spacetime metric, in the same way as the ordinary Euclidean distance computed using the Pythagorean formula is the dot product of the spatial position vector with itself.

    The energy-momentum 4-vector works the same way; in fact, *any* 4-vector in spacetime works the same way (just as we can compute the dot product of any ordinary 3-vector in Euclidean space the same way we did above for the position vector). That's why there's the minus sign in front of E^2. The sign convention (minus sign in front of E^2, instead of in front of the p^2 components) is something we already talked about, but I'll go into it a bit more in the answer to your other question below.

    The short answer is that, as you've written it, the second form is right and the first is not. However, there's a deeper issue here which is worth going into.

    The proper way of writing a 4-vector, the things we've been talking about up to now, is with the index "upstairs", as you wrote it. So the ordinary "position vector" would be

    [tex]x^{\mu} = (x, y, z, t)[/tex]

    Notice that there is no minus sign in front of the t. Similarly, the energy-momentum 4-vector would be

    [tex]p^{\mu} = (p^x, p^y, p^z, \frac{E}{c})[/tex]

    with no minus sign in front of the E. (Note also that I wrote the x, y, z on the p components "upstairs", not "downstairs" as you wrote them. We'll come back to that.)

    You will also, however, see objects written with the index "downstairs". For example, you might see something like this:

    [tex]p_{\mu} = (p_x, p_y, p_z, - \frac{E}{c})[/tex]

    with a minus sign in front of the E. What's going on here?

    The answer is that the object with the "downstairs" index is not a vector; it's a different kind of object, usually called a "1-form" or "covector". You can read some about it here:


    We don't need to go into a lot of detail about 1-forms; the key point is that, as long as we have a metric (which we do here), there is a 1-to-1 mapping between 1-forms and vectors, using the metric, which is written this way:

    [tex]p_{\mu} = \eta_{\mu \nu} p^{\nu}[/tex]

    That [itex]\eta_{\mu \nu}[/itex] is the "metric tensor", which for our purposes here you can just think of as a 2 x 2 matrix with (1, 1, 1, -1) along the diagonal and 0 everywhere else, using your professor's sign convention. The metric tensor is also what we use to form the dot product of vectors, so we can write the energy-momentum relation as the dot product of the 4-momentum vector with itself thus:

    [tex]\eta_{\mu \nu} p^{\mu} p^{\nu} = p^1 p^1 + p^2 p^2 + p^3 p^3 - p^0 p^0 = ( p^x )^2 + ( p^y )^2 + ( p^z )^2 - \left( \frac{E}{c} \right)^2 = - m^2 c^2[/tex]

    where we have used a very useful convention called the "Einstein summation convention", in which any index that is repeated (i.e., it appears both "upstairs" and "downstairs") is summed over, with values (1, 2, 3, 0) corresponding to the (x, y, z, t) components of the vector. I used the same convention in writing the mapping from vectors to 1-forms, but since the metric tensor is diagonal, the sum for each component collapses to only one term, and we have

    p_1 = \eta_{11} p^1 = p^x = p_x \\
    p_2 = \eta_{22} p^2 = p^y = p_y \\
    p_3 = \eta_{33} p^3 = p^z = p_z \\
    p_0 = \eta_{00} p^0 = - \frac{E}{c}

    That's where the minus sign comes from in the 1-form. Also, as you can see, the spatial momentum components are the same for the vector and the 1-form, so it doesn't really matter whether we write them "upstairs" or "downstairs" if we are using your professor's sign convention. (As an exercise, though, you might want to go back and re-write all these formulas using my sign convention. The first thing to re-write is the metric tensor: what does it look like with my sign convention?)

    This is a lot to digest so I'll stop now. Please feel free to ask further questions when you've looked it over.
    Last edited: Dec 5, 2012
  16. Dec 6, 2012 #15
    I don't mean to be rude, but being a touch more humble will help you along your academic road..
    I see that you're using modern physics in your class, that book barely goes into any depths with SR.
    To properly master SR you need to learn about covariant and contravariant transformations, and thus basically tensor analysis.
    Transforming between inertial frames becomes easy once you treat things with four vectors and one forms, just drop a lambda matrix in front of your vector and poof, transformed.
    The things bout four vectors is that they are the same geometrical object in every inertial frame, while three vectors are not. Understand this, the components on a vector can change, but a four vector is the same in every single inertial frame.
    Last edited: Dec 6, 2012
  17. Dec 7, 2012 #16
    I will definitely take this on my knowledge. Thank you. At the moment i am still studying the anwser above, so i ll keep in touch in case i get more questions.
  18. Dec 7, 2012 #17
    The central idea for SR is that space and time are bounded together into one mathematical model, the space time. In the same sense in classical mechanics that your y and your x mean nothing to nature, your t and your x mean nothing to nature, they are simply different viewpoints of a single space-time.(Or more mathematically, simply different inertial frames that we choose to work in)
    We use tensors because they embody this concept, the components of a tensor transform with a matrix that is the inverse of the matrix that their bases transform through, hence when you write a tensor out in a basis the coordinate transform matrices multiply and become the identity matrix, the "1" in multilinear algebra.
    Many would put this in more mathematical terms, and say that the invariance of the space-time interval is the center piece of SR, this is true in a sense. The convention we choose for it is not important, whether you use (+---) or (-+++) the important thing is that this be kept invariant of which inertial frame we are in.
    Even more fundamentally, the inner product ds^2=-dt^2+dS^2 is actually just the action of the metric tensor on the same four vector twice, measuring the "length" of that four vector.
    This is more fundamental because when you move on to GR, you learn that the metric is different for different systems, depending on the energy and momentum flow in that region of space time. This leads to many other strange qualities of realistic mechanics:

    -Because the metric is in general, position dependent, we can no longer think of vectors as arrows spanning a length in space time, but rather an entity that exists at each point of space time.

    -This leads to the fact that relative velocities are meaningless in GR, if you compare two four velocities, you need to move them to one point in space time and compare them. However as it turns out in curved space (where the metric is a function of space and time), the way you "slide" the two vectors affects them and hence there is no way to compare two vectors that inhabit different points in space time.

    -There are no lorentz frames, due to the fact that gravity is not something that can be isolated, in the sense that you cannot shield a particle from gravity, there truly exists no inertial frames in curved space time, everyone is in free fall.

    p.s. Yes, Physics does take pleasure in making your integrals and PDEs harder and nastier. If I know one thing it's that.
    Last edited: Dec 7, 2012
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