Physical laws and tensor formulations of them

[McLaren Rulez]

Hi,

I am reading up on special relativity and I'm having some trouble understanding how tensors fit into the picture. Its my first contact with these concepts so please forgive me for being very muddled. My main problem is understanding how to see whether a physical law is compatible with Lorentz transformation.

Wikipedia has an article here (http://en.wikipedia.org/wiki/Lorentz_invariant) where they say that any equation made of Lorentz covariant quantities will transform correctly when we change to a new frame of reference. It says that this is because of the fact that if all the components of a tensor vanish in one frame, they vanish in every frame. But which tensor are we talking about here? How did a tensor emerge from one equation?

Also, what does the phrase "equation made of Lorentz covariant quantities" mean. For example, if I take some equation like

$$ax_1{}+bV_4{}+cF_2{}=0$$

where x, V and F are the usual four vectors, then is this an equation that transforms correctly (never mind the fact that its rubbish)? And if so, why does it transform correctly?

Thank you.

 

[bcrowell]

Wikipedia has an article here (http://en.wikipedia.org/wiki/Lorentz_invariant) where they say that any equation made of Lorentz covariant quantities will transform correctly when we change to a new frame of reference. It says that this is because of the fact that if all the components of a tensor vanish in one frame, they vanish in every frame. But which tensor are we talking about here? How did a tensor emerge from one equation?

Probably what they have in mind is the tensor defined as the difference between the tensor on the right-hand side and the tensor on the left-hand side of the equation.

Also, what does the phrase "equation made of Lorentz covariant quantities" mean. For example, if I take some equation like

$$ax_1{}+bV_4{}+cF_2{}=0$$

where x, V and F are the usual four vectors, then is this an equation that transforms correctly (never mind the fact that its rubbish)? And if so, why does it transform correctly?

Do the 1, 4, and 2 refer to specific components of these four-vectors? If so, then this isn't an equation made of Lorentz covariant quantities. An equation made of Lorentz-covariant quantities can only have dummy indices in it, e.g., $ax_k+bV_k+cF_k=0$.

All of this stuff can be understood by analogy with ordinary three-vectors. $F=ma$ is a valid vector equation, so if it's valid for one coordinate system, it's also valid if you rotate that coordinate system. Writing it in tensor-gymnastics notation, it becomes $F_\mu=ma_\mu$, where $\mu=x,y,z$. An equation like $F_x+ma_y=0$ isn't a valid vector equation. Even if it happened to be true in one coordinate system, it would be false if you rotated.

 

[Mentz114]

To be covariant means that an object transforms so that any scalars formed by contraction will invariant.

Scalars transform without change, 4-vectors transform so their magnitude ( a scalar) is invariant.

This equation is covariant
$$J^\mu = {F^{\mu\nu}}_{;\nu}$$
because it makes a 4-vector from the contraction of a tensor. It is tautologous to say that tensors transform covariantly because a tensor is defined as an object that transforms covariantly.

Writing an expression purely in terms of scalars, 4-vectors and tensors ensures covariance.

 

[McLaren Rulez]

Thank you for the replies.

Probably what they have in mind is the tensor defined as the difference between the tensor on the right-hand side and the tensor on the left-hand side of the equation.

$F=ma$ is a valid vector equation, so if it's valid for one coordinate system, it's also valid if you rotate that coordinate system. Writing it in tensor-gymnastics notation, it becomes $F_\mu=ma_\mu$, where $\mu=x,y,z$

Okay so in this case, is the tensor actually a four vector with the following components, which are, of course, zero in every frame.
$$F_\mu{}-ma_\mu$$
Is that right?

To be covariant means that an object transforms so that any scalars formed by contraction will invariant.

Scalars transform without change, 4-vectors transform so their magnitude ( a scalar) is invariant.

This equation is covariant
$$J^\mu = {F^{\mu\nu}}_{;\nu}$$
because it makes a 4-vector from the contraction of a tensor. It is tautologous to say that tensors transform covariantly because a tensor is defined as an object that transforms covariantly.

Writing an expression purely in terms of scalars, 4-vectors and tensors ensures covariance.

Sorry, I'm not familiar with the contraction of tensors. From what I understand, contraction is when you multiply two tensors and set two free indices to be equal to get a tensor of lower rank. I guess you are going from a four vector to a scalar (its magnitude) but how did you use contraction?

I also understand how four vectors have a constant magnitude in all frames. But I don't follow beyond that. Basically, why would writing an equation in terms of four vectors ensure that it transforms correctly.

Thank you, once again. The picture in my head is definitely getting clearer but there is a lot to learn for me.

 

[Mentz114]

Thank you for the replies.
Sorry, I'm not familiar with the contraction of tensors. From what I understand, contraction is when you multiply two tensors and set two free indices to be equal to get a tensor of lower rank. I guess you are going from a four vector to a scalar (its magnitude) but how did you use contraction?

This expression ${F^{\mu\nu}}_{;\nu}$ is a sum ${F^{\mu 0}}_{;0}+{F^{\mu 1}}_{;1}+{F^{\mu 2}}_{;2}+{F^{\mu 3}}_{;3}$. This is a rank-3 tensor contracted over 2 indexes leaving one, which is a rank-1 tensor ( 4-vector). The ';' indicates covariant differentiation which isn't important in this context, but I thought you might like to know.

But I don't follow beyond that. Basically, why would writing an equation in terms of four vectors ensure that it transforms correctly.

I'm sure there is a proper mathematical proof of this, but I don't know it.

 

[McLaren Rulez]

How about this equation? $$(id/dt+icd/dx)\psi=0$$
The d/dt and d/dx are partial derivatives and the psi is a wavefunction. I believe this is the massless Dirac equation in two dimensions. Can anyone show me how this can be written in terms of the dummy indices in four dimensions making it Lorentz covariant?

Thank you.

 

[Mentz114]

How about this equation? $$(id/dt+icd/dx)\psi=0$$
The d/dt and d/dx are partial derivatives and the psi is a wavefunction. I believe this is the massless Dirac equation in two dimensions. Can anyone show me how this can be written in terms of the dummy indices in four dimensions making it Lorentz covariant?

Thank you.

See chapter XI in Dirac, 'Principles of Quantum Mechanics'. The momentum 4-vector is introduced and the 4-momentum operator
$$p_\mu=\frac{i\partial}{\partial x^\mu}$$
The time component of the momentum $p_0$ is the energy (working in units where c=1).

 

[McLaren Rulez]

Okay looks like I need to get myself a decent textbook for this. Thank you for the help so far, and if there are any recommendations for good texts, do let me know. Thanks!