Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why the Minkowski inner product is not positive-definite?

  1. Dec 9, 2013 #1
    Hi,
    I know these questions must sound ridiculous and I apologize, I'm a newbie. My textbook says that the inner product of the momentum four-vector is
    P[itex]\bullet[/itex]P=P[itex]\bullet[/itex]P - E[itex]^{2}[/itex]/c[itex]^{2}[/itex]=-m[itex]^{2}[/itex]*c[itex]^{2}[/itex]
    So my silly questions are: 1) where did the - E[itex]^{2}[/itex]/c[itex]^{2}[/itex] term come from? 2) I know I'm being a pain but what does the term 'Minkowski' mean in this context? I only have a passing familiarity with the Minkowski metric which I mentally make sense of using spacetime intervals so I'm not sure why the inner product of the momentum four-vector is being called the Minkowski inner product. 3) For the sake of intuition, what does the -m[itex]^{2}[/itex]*c[itex]^{2}[/itex] result 'mean'? (I'm particularly confused over the fact it's negative).
    Thanks for your time.
     
  2. jcsd
  3. Dec 9, 2013 #2
    The energy of a particle is defined as the "time-component" of the 4-momentum, vector. In non-geometrized units, that component of the 4-momentum is E/c. The -(E^2)/(c^2) term comes from multiplying out the 4-momentum with itself: p*p = -p0*p0 + p1*p1 + p2*p2 + p3*p3. Of course, p0 = E/c, so p*p = -(E/c)^2+....

    The Minkowski metric refers to a flat metric of signature (1,3). The inner product of vectors in Minkowski space is defined with the Minkowski metric, that is, for any two vectors v and w, we have inner product v*w = -v0w0 + v1w1 + v2w2 + v3w3.
     
  4. Dec 9, 2013 #3

    K^2

    User Avatar
    Science Advisor

    A dot product, or inner product, cannot be taken between two vectors from the same space. Typically, we talk about a vector space and a dual vector space. The reason this doesn't very frequently come up outside of relativity or higher mathematics is because a dual to Eucledian vector is an identical Eucledian vector. So when taking Eucledian dot-products, we can forget about this detail.

    Not so in general. The simplest example you are probably familiar with is matrix representation of vectors. I can write down a 3D vector as a 1x3 matrix containing x, y, and z on corresponding columns. Now, if I have another vector, it's another row matrix, and I cannot multiply two row matrices together. However, the dual to row matrix vectors are the column matrices. If I now write down the second vector as a column (x, y, z)T, I can multiply the two together and you will get the ordinary dot product of two vectors.

    The other example is Minkowski space. Typically, we denote vectors with an upper subscript and dual vectors with a lower subscript. (Or vice versa. In L2, dual of a dual is isomorphic to original space. So it doesn't matter which is the original space and which is dual.) So in order to take a dot product between 4-vectors a and b, we need the vector ##a^{\mu}## and the dual of b ##b_{\mu}##. The Greek sub/super scripts denote abstract index notation. In other words, we don't mean any particular basis, but only use the indices to denote which is vector and which is dual.

    With matrices, to go from one space to another you just took a transpose. The generalization of the concept is metric tensor. It is a rank 2 tensor written ##g_{\mu\nu}##. In Minkowski space, this is the Minkowski metric tensor.

    In a particular coordinate system, we can represent all of this, again, with matrices. A vector ##a^{\mu}## can be represented by a row matrix (at, ax, ay, az). Now the superscripts denote that these are specific components of the vector in a chosen coordinate system. Suppose that we also have a vector ##b^{\mu}## with similar representation. We want to get the dual of ##b^{\mu}## so that we can take a dot product. In abstract notation, the product with tensor is written as ##g_{\mu\nu}b^{\nu}##. The actual choice of letters is arbitrary. What's important is that like indices are contracted. In matrix notation, this means summation over indices, which we are going to write out as matrix multiplication. Specifically, the matrix representation of ##b_{\mu}## is bi = gijbj. Here, the fact that j index is repeated means we are going to take a sum over all indices. For example, bt = gttbt + gtxbx + gtyby + gtzbz.

    So it's time to talk about what the gij matrix looks like. In Eucledian space, it's an identity matrix. So bi = bi. And that means you can take a dot product by just taking the sum of products of corresponding terms. In Minkowski metric, that matrix representation is still diagonal, so all terms like gxy are zero. But diagonal terms have different signs. Specifically, gtt = -1, and gxx, gyy, and gzz = +1. (This is consistent with your question and referred to as (-+++) signature. Alternatively, there is (+---) signature where signs are flipped.)

    So now, the dot product of vectors a and b follows naturally, giving you ##a^{\mu}b_{\mu}## = axbx + ayby + azbz - atbt. Here the spacial part is just an ordinary dot product, so that this can also be written as ##a\cdot b## - atbt. Note that if I use vector (ct, x, y, z) and take a dot product with itself using this prescription, I get the standard space-time interval expression x² + y² + z² - c²t².

    Finally, representation of the momentum 4-vector is ##p^{\mu}## = (E/c, px, py, pz), which leads to the dot product ##p^{\mu}p_{\mu}## from your post.

    Whether the result is +mc² or -mc² is up to the choice of signature above. It's just a matter of convention, and in a given convention, there is no other significance.
     
  5. Dec 9, 2013 #4
    Those were great answers--K^2 's answer even managed to help me get over my indice-phobia. Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why the Minkowski inner product is not positive-definite?
Loading...